снартев 3

THE ELECTROMAGNETIC FIELD

The rotating armatures of every generator and every motor in this age of electricity are steadily proclaiming the truth of the relativity theory to all who have ears to hear.

LEIGH PAGE (1941)

§3.1. THE LORENTZ FORCE AND THE ELECTROMAGNETIC FIELD TENSOR

At the opposite extreme from an impulsive change of momentum in a collision (the last topic of Chapter 2) is the gradual change in the momentum of a charged particle under the action of electric and magnetic forces (the topic treated here).

Let electric and magnetic fields act on a system of charged particles. The accelerations of the particles reveal the electric and magnetic field strengths. In other words, the Lorentz force law, plus measurements on the components of acceleration of test particles, can be viewed as defining the components of the electric and magnetic fields. Once the field components are known from the accelerations of a few test particles, they can be used to predict the accelerations of other test particles (Box 3.1). Thus the Lorentz force law does double service (1) as definer of fields and (2) as predicter of motions.
Here and elsewhere in science, as stressed not least by Henri Poincaré, that view is out of date which used to say, "Define your terms before you proceed." All the laws and theories of physics, including the Lorentz force law, have this deep and subtle character, that they both define the concepts they use (here B and

E

) and make statements about these concepts. Contrariwise, the absence of some body of theory, law, and principle deprives one of the means properly to define or even to use concepts. Any forward step in human knowledge is truly creative in this sense: that theory, concept, law, and method of measurement-forever inseparable-are born into the world in union.

Lorentz force as definer of fields and predicter of motions

Box 3.1 LORENTZ FORCE LAW AS BOTH DEFINER OF FIELDS AND PREDICTER OF MOTIONS

How one goes about determining the components of the field from measurements of accelerations is not different in principle for electromagnetism and for gravitation. Compare the equations in the two cases:

\begin{matrix} (1) & \frac{d^{2} x^{α}}{d τ^{2}} = \frac{e}{m} F^{α}_{β} u^{β} in a Lorentz frame, \end{matrix}

and

\frac{D^{2} ξ^{α}}{d τ^{2}} = - R^{α}_{β γ δ} u^{β} ξ^{γ} u^{δ}

in any coordinate system.

To make explicit the simpler procedure for electromagnetism will indicate in broad outline how one similarly determines all the components of

R^{α}_{β γ δ}

for gravity. Begin by asking how many test particles one needs to determine the three components of

B

and the three components of

E

in the neighborhood under study. For one particle, three components of acceleration are measurable; for a second particle, three more. Enough? No! The information from the one duplicates in part the information from the other. The proof? Whatever the state of motion of the first test particle, pick one's Lorentz frame to be moving the same way. Having zero velocity in this frame, the particle has a zero response to any magnetic field. The electric field alone acts on the particle. The three components of its acceleration give directly the three components

E_{x}, E_{y}, E_{z}

of the electric field. The second test particle cannot be at rest if it is to do more than duplicate the information provided by the first test particle. Orient the

x

-axis of the frame
of reference parallel to the direction of motion of this second particle, which will then respond to and measure the components

B_{y}

and

B_{z}

of the magnetic field. Not so

B_{x}

! The acceleration in the

x

-direction merely remeasures the already once measured

E_{x}

. To evaluate

B_{x}

, a third test particle is required, but it then gives duplicate information about the other field components. The alternative? Use all

N

particles simultaneously and on the same democratic footing, both in the evaluation of the six

F_{α β}

and in the testing of the Lorentz force, by applying the method of least squares. Thus, write the discrepancy between predicted and observed acceleration of the

K

th particle in the form

\begin{matrix} (3) & {\dot{u}}_{α}^{K} - \frac{e}{m} F_{α β} u^{β, K} = δ a_{α}^{K} \end{matrix}

Take the squared magnitude of this discrepancy and sum over all the particles

\begin{matrix} (4) & S = \sum_{k} η^{α β} δ a_{α}^{K} δ a^{K}_{β} . \end{matrix}

In this expression, everything is regarded as known except the six

F_{α β}

. Minimize with respect to these six unknowns. In this way, arrive at six equations for the components of

B

and

E

. These equations once solved, one goes back to (3) to test the Lorentz force law.
The

6 \times 6

determinant of the coefficients in the equation for the

F_{α β}

automatically vanishes when there are only two test particles. The same line of reasoning permits one to determine the minimum number of test particles required to determine all the components of the Riemann curvature tensor.

The Lorentz force law, written in familiar three-dimensional notation, with

E =

electric field,

B =

magnetic field,

v =

ordinary velocity of particle,

p =

momentum of particle,

e =

charge of particle, reads

\begin{matrix} (3.1) & (d p / d t) = e (E + v \times B) \end{matrix}

Useful though this version of the equation may be, it is far from the geometric spirit of Einstein. A fully geometric equation will involve the test particle's energy-momentum 4 -vector,

p

, not just the spatial part

p

as measured in a specific Lorentz frame; and it will ask for the rate of change of momentum not as measured by a specific Lorentz observer

(d / d t)

, but as measured by the only clock present à priori in the problem: the test particle's own clock

(d / d τ)

. Thus, the lefthand side of a fully geometric equation will read

d p / d τ =

The righthand side, the Lorentz 4 -force, must also be a frame-independent object. It will be linear in the particle's 4 -velocity

u

, since the frame-dependent expression

\begin{matrix} (3.2a) & \frac{d p}{d τ} = \frac{1}{\sqrt{1 - v^{2}}} \frac{d p}{d t} = \frac{e}{\sqrt{1 - v^{2}}} (E + v \times B) = e (u^{0} E + u \times B) \end{matrix}

is linear in the components of

u

. Consequently, there must be a linear machine named Faraday, or

F

, or "electromagnetic field tensor," with a slot into which one inserts the 4 -velocity of a test particle. The output of this machine, multiplied by the particle's charge, must be the electromagnetic 4 -force that it feels:

\begin{matrix} (3.3) & d p / d τ = e F (u) \end{matrix}

By comparing this geometric equation with the original Lorentz force law (equation 3.2 a ), and with the companion energy-change law

\begin{matrix} (3.2b) & \frac{d p^{0}}{d τ} = \frac{1}{\sqrt{1 - v^{2}}} \frac{d E}{d t} = \frac{1}{\sqrt{1 - v^{2}}} e E \cdot v = e E \cdot u \end{matrix}

one can read off the components of

F

in a specific Lorentz frame. The components of

d p / d τ

are

d p^{α} / d τ

, and the components of

e F (u)

can be written (definition of

F^{α}_{β}!

)

e F^{α}_{β} u^{β}

. Consequently

\begin{matrix} (3.4) & d p^{α} / d τ = e F^{α}_{β} u^{β} \end{matrix}

must reduce to equations

(3.2 a, b)

. Indeed it does if one makes the identification

\begin{matrix} (3.5) & \begin{array}{r} β = 0 \\ α = 1 \\ α = 0 \\ ‖ F^{α}_{β} ‖ = \begin{array}{l} α = 2 \\ α = 1 \\ α \\ α \\ α \end{array} ‖^{0} \end{array} ‖ \begin{array}{cccc} E_{x} & E_{y} & E_{z} \\ E_{x} & 0 & B_{z} & - B_{y} \\ E_{y} & - B_{z} & 0 & B_{x} \\ E_{z} & B_{y} & - B_{x} & 0 \end{array} ‖ . \end{matrix}

The three-dimensional version of the Lorentz force law

Electromagnetic field tensor defined

Geometrical version of Lorentz force law

Components of electromagnetic field tensor

More often seen in the literature are the "covariant components," obtained by lowering an index with the metric components:

\begin{matrix} (3.6) & F_{α β} = η_{α γ} F_{β}^{γ} \\ (3.7) & ‖ F_{α β} ‖ = ‖ \begin{array}{cccc} 0 & - E_{x} & - E_{y} & - E_{z} \\ E_{r} & 0 & B_{z} & - B_{y} \\ E_{y} & - B_{z} & 0 & B_{x} \\ E_{z} & B_{y} & - B_{x} & 0 \end{array} ‖ . \end{matrix}

This matrix equation demonstrates the unity of the electric and magnetic fields. Neither one by itself,

E

B

, is a frame-independent, geometric entity. But merged together into a single entity,

F = F a r a d a y

, they acquire a meaning and significance that transcends coordinates and reference frames.

EXERCISE

Exercise 3.1.

Derive equations (3.5) and (3.7) for the components of Faraday by comparing (3.4) with (3.2a,b), and by using definition (3.6).

§3.2. TENSORS IN ALL GENERALITY

Examples of tensors
A digression is in order. Now on the scene are several different tensors: the metric tensor

g

(§2.4), the Riemann curvature tensor Riemann (§1.6), the electromagnetic field tensor Faraday ($3.1). Each has been defined as a linear machine with input slots for vectors, and with an output that is either a real number, e.g.,

g (u, v)

, or a vector, e.g., Riemann (u,v,w) and Faraday (u).

Should one make a distinction between tensors whose outputs are scalars, and tensors whose outputs are vectors? No! A tensor whose output is a vector can be reinterpreted trivially as one whose output is a scalar. Take, for example, Fara

d a y = F

. Add a new slot for the insertion of an arbitrary 1 -form

σ

, and gears and wheels that guarantee the output

\begin{matrix} (3.8) & F (σ, u) = ⟨ σ, F (u) ⟩ = real number. \end{matrix}

Then permit the user to choose whether he inserts only a vector, and gets out the vector

F (\dots, u) = F (u)

, or whether he inserts a form and a vector, and gets out the number

F (σ, u)

. The same machine will do both jobs. Moreover, in terms of components in a given Lorentz frame, both jobs are achieved very simply:

F (\dots, u)

is a vector with components

F^{α}_{β} u^{β}

;

F (σ, u)

is the number

⟨ σ, F (\dots, u) ⟩ = σ_{α} F^{α}_{β} u^{β}

By analogy, one defines the most general tensor

H

and its rank

(\binom{n}{m})

as follows:

H

is a linear machine with

n

input slots for

n 1

-forms, and

m

input slots for

m

vectors; given the requested input, it puts out a real number denoted

\begin{matrix} (3.10) & H (\underset{n 1-forms}{\underset{⏟}{σ, λ, \dots, β}}, \underset{m vectors}{\underset{⏟}{, u, v, \dots, w)}} \end{matrix}

For most tensors the output changes when two input vectors are interchanged,

\begin{matrix} (3.11) & Riemann (σ, u, v, w) \neq Riemann (σ, v, u, w) \end{matrix}

or when two input 1 -forms are interchanged.
Choose a specific tensor

S

, of rank

(\binom{2}{1})

for explicitness. Into the slots of

S

, insert the basis vectors and 1 -forms of a specific Lorentz coordinate frame. The output is a "component of

S

in that frame":

\begin{matrix} (3.12) & S^{α β}_{γ} \equiv S (ω^{α}, ω^{β}, e_{γ}) . \end{matrix}

This defines components. Knowing the components in a specific frame, one can easily calculate the output produced from any input forms and vectors:

\begin{aligned} (3.13) & S (σ, ρ, v) & = S (σ_{α} ω^{α}, ρ_{β} ω^{β}, v^{γ} e_{γ}) = σ_{α} ρ_{β} v^{γ} S (ω^{α}, ω^{β}, e_{γ}) \\ = S^{α β}_{γ} σ_{α} ρ_{β} v^{γ} . \end{aligned}

And knowing the components of

S

in one Lorentz frame (unprimed), plus the Lorentz transformation matrices

‖ Λ^{α^{'}}_{β} ‖

and

‖ Λ^{β}_{α^{'}} ‖

which link that frame with another (primed), one can calculate the components in the new (primed) frame. As shown in exercise 3.2 , one need only apply a matrix to each index of

S

, lining up the matrix indices in the logical manner

\begin{matrix} (3.14) & S^{μ^{'} ν^{'}}_{λ^{'}} = S^{α β}_{γ} Λ^{μ^{'}}_{α} Λ^{v^{'}}_{β} Λ^{γ}_{λ^{'}} \end{matrix}

A slight change of the internal gears and wheels inside the tensor enables one of its 1 -form slots to accept a vector. All that is necessary is a mechanism to convert an input vector

n

into its corresponding 1 -form

\tilde{n}

and then to put that 1 -form into the old machinery. Thus, denoting the modified tensor by the same symbol

S

as was used for the original tensor, one demands

\begin{matrix} (3.15) & S (σ, n, v) = S (σ, \tilde{n}, v) \end{matrix}

or, in component notation

\begin{matrix} (') & S^{α}_{β γ} σ_{α} n^{β} v^{γ} = S^{α β}_{γ} σ_{α} n_{β} v^{γ} . \end{matrix}

This is achieved if one raises and lowers the indices of

S

using the components of the metric:

\begin{matrix} (3.16) & S^{α}_{β γ} = η_{β μ} S^{α μ}_{γ}, S^{α μ}_{γ} = η^{μ β} S^{α}_{β γ} . \end{matrix}

(See exercise 3.3 below.) By using the same symbol

S

for the original tensor and

Definition of tensor as linear machine that converts vectors and 1 -forms into numbers

Components of a tensor

Tensor's machine action expressed in terms of components

Lorentz transformation of components

Modifying a tensor to accept either a vector or a 1 -form into each slot

Raising and lowering indices
the modified tensor, one allows each slot to accept either a 1 -form or a vector, so one loses sight of whether

S

is a

(\binom{2}{1})

tensor, or a

(\binom{1}{2})

tensor, or a

(\binom{3}{0})

tensor, or a

(\binom{0}{3})

tensor; one only distinguishes its total rank, 3. Terminology: an "upstairs index" is called "contravariant"; a "downstairs" index is called "covariant." Thus in

S^{α}_{β γ}

, "

α

" is a contravariant index, while "

β

" and "

γ

" are covariant indices.

Because tensors are nothing but functions, they can be added (if they have the same rank!) and multiplied by numbers in the usual way: the output of the rank-3 tensor

a S + b Q

, when vectors

u, v, w

are put in, is

\begin{matrix} (3.17) & (a S + b Q) (u, v, w) \equiv a S (u, v, w) + b Q (u, v, w) \end{matrix}

Several other important operations on tensors are explored in the following exercises. They and the results of the exercises will be used freely in the material that follows.

EXERCISES

Exercise 3.2. TRANSFORMATION LAW FOR COMPONENTS OF A TENSOR

From the transformation laws for components of vectors and 1 -forms, derive the transformation law (3.14).

Exercise 3.3. RAISING AND LOWERING INDICES

Derive equations (3.16) from equation (3.15') plus the law

n_{α} = η_{α β} n^{β}

for getting the components of the 1 -form

\tilde{n}

from the components of its corresponding vector

n

Exercise 3.4. TENSOR PRODUCT

Given any two vectors

u

and

v

, one defines the second-rank tensor

u \otimes v

("tensor product of

u

with

v

") to be a machine, with two input slots, whose output is the number

\begin{matrix} (3.18) & (u \otimes v) (σ, λ) = ⟨ σ, u ⟩ ⟨ λ, v ⟩ \end{matrix}

when 1-forms

σ

and

λ

are inserted. Show that the components

T T = u \otimes v

are the products of the components of

u

and

v

\begin{matrix} (3.19) & T^{α β} = u^{α} v^{β}, T_{α}^{β} = u_{α} v^{β}, T_{α β} = u_{α} v_{β} . \end{matrix}

Extend the definition to several vectors and forms,

\begin{matrix} (3.20) & (u \otimes v \otimes β \otimes w) (σ, λ, n, ζ) = ⟨ σ, u ⟩ ⟨ λ, v ⟩ ⟨ β, n ⟩ ⟨ ζ, w ⟩, \end{matrix}

and show that the product rule for components still holds:

\begin{matrix} (3.21) & S = u \otimes v \otimes β \otimes w has components \\ S_{λ^{μ}}^{μ ν} = u^{μ} v^{ν} β_{λ} w^{ζ} . \end{matrix}

Exercise 3.5. BASIS TENSORS

Show that a tensor

M

with components

M^{α β} γ^{δ}

in a given Lorentz frame can be reconstructed from its components and from the basis 1 -forms and vectors of that frame as follows:

\begin{matrix} (3.22) & M = M^{α β} γ^{δ} e_{α} \otimes e_{β} \otimes ω^{γ} \otimes e_{δ} \end{matrix}

(For a special case of this, see Box 3.2.)

Box 3.2 THE METRIC IN DIFFERENT LANGUAGES

A. Geometric Language

g

is a linear, symmetric machine with two slots for insertion of vectors. When vectors

u

and

v

are inserted, the output of

g

is their scalar product:

g (u, v) = u \cdot v

B. Component Language

η_{μ ν}

are the metric components. They are used to calculate the scalar product of two vectors from components in a specific Lorentz frame:

u \cdot v = η_{μ ν} u^{μ} v^{ν}

C. Coordinate-Based Geometric Language

The metric

g

can be written, in terms of basis 1-forms of a specific Lorentz frame, as

g = η_{μ ν} ω^{μ} \otimes ω^{ν} = η_{μ ν} d x^{μ} \otimes d x^{ν}

[see equations (2.18) and (3.22)].

D. Connection to the Elementary Concept of Line Element

Box 2.3 demonstrated the correspondence between the gradient

d f

of a function, and the elementary concept

d f

of a differential change of

f

in some unspecified direction. There is a similar correspondence between the metric, written as

η_{μ ν} d x^{μ}

\otimes d x^{ν}

, and the elementary concept of "line element," written as

d s^{2} = η_{μ ν} d x^{μ} d x^{ν}

. This elementary line element, as expounded in many special relativity texts, represents the squared length of the displacement "

d x

" " in an unspecified direction. The metric

η_{μ ν} d x^{μ} \otimes d x^{ν}

does the same. Pick a specific infinitesimal displacement vector

ξ

, and insert it into the slots of

η_{μ ν} d x^{μ} \otimes d x^{ν}

. The output will be

ξ^{2} = η_{μ ν} ξ^{μ} ξ^{ν}

, the squared length of the displacement. Before

ξ

is inserted,

η_{μ ν} d x^{μ} \otimes d x^{ν}

has the potential to tell the squared length of any vector; the insertion of

ξ

converts potentiality into actuality: the numerical value of

ξ^{2}

Because the metric

η_{μ ν} d x^{μ} \otimes d x^{ν}

and the line element

d s^{2} = η_{μ ν} d x^{μ} d x^{ν}

perform this same function of representing the squared length of an unspecified infinitesimal displacement, there is no conceptual distinction between them. One sometimes uses the symbols

d s^{2}

to denote the metric; one sometimes gets pressed and writes it as

d s^{2} = η_{μ ν} d x^{μ} d x^{ν}

, omitting the "

\otimes

"; and one sometimes even gets so pressed as to use nonbold characters, so that no notational distinction remains at all between metric and elementary line element:

g = d s^{2} = d s^{2} = η_{μ ν} d x^{μ} d x^{ν}

Exercise 3.6. Faraday MACHINERY AT WORK

An observer with 4 -velocity

u

picks out three directions in spacetime that are orthogonal and purely spatial (no time part) as seen in his frame. Let

e_{\hat{1}}, e_{\hat{2}}, e_{\hat{3}}

be unit vectors in those directions and let them be oriented in a righthanded way

(e_{1} \cdot e_{2} \times e_{3} = + 1

in three-dimensional language). Why do the following relations hold?

e_{j} \cdot u = 0, e_{j} \cdot e_{k} = δ_{j k}

What vectors are to be inserted in the two slots of the electromagnetic field tensor Faraday if one wants to get out the electric field along

e_{j}

as measured by this observer? What vectors must be inserted to get the magnetic field he measures along

e_{j}

§3.3. THREE-PLUS-ONE VIEW VERSUS GEOMETRIC VIEW

The power of the geometric view of physics

Example of electromagnetism

Transformation law for electric and magnetic fields

Great computational and conceptual power resides in Einstein's geometric view of physics. Ideas that seem complex when viewed in the everyday "space-plus-time" or "

3 + 1

" manner become elegant and simple when viewed as relations between geometric objects in four-dimensional spacetime. Derivations that are difficult in

3 + 1

language simplify in geometric language.

The electromagnetic field is a good example. In geometric language, it is described by a second-rank, antisymmetric tensor ("2-form")

F

, which requires no coordinates for its definition. This tensor produces a 4 -force on any charged particle given by

d p / d τ = e F (u) .

It is all so simple!
By contrast, consider the "

3 + 1

" viewpoint. In a given Lorentz frame, there is an electric field

E

and a magnetic field

B

. They push on a particle in accordance with

d p / d t = e (E + v \times B)

But the values of

p, E, v

, and

B

all change when one passes from the given Lorentz frame to a new one. For example, the electric and magnetic fields viewed from a rocket ship ("barred" frame) are related to those viewed in the laboratory ("unbarred" frame) by

\begin{matrix} (3.23) & \begin{array}{ll} {\overset{―}{E}}_{‖} = E_{‖}, & {\overset{―}{E}}_{⊥} = \frac{1}{\sqrt{1 - β^{2}}} (E_{⊥} + β \times B_{⊥}), \\ {\overset{―}{B}}_{‖} = B_{‖}, & {\overset{―}{B}}_{⊥} = \frac{1}{\sqrt{1 - β^{2}}} (B_{⊥} - β \times E_{⊥}) . \end{array} \end{matrix}

(Here "Il" means component along direction of rocket's motion; "

⊥

" means perpendicular component; and

β^{j} = d x^{j}

rocket

/ d t

is the rocket's ordinary velocity.) The analogous transformation laws for the particle's momentum

p

and ordinary velocity

v

, and for the coordinate time

t

, all conspire-as if by magic, it seems, from the

3 + 1

viewpoint-to maintain the validity of the Lorentz force law in all frames.

Not only is the geometric view far simpler than the

3 + 1

view, it can even derive the

3 + 1

equations with greater ease than can the

3 + 1

view itself. Consider, for example, the transformation law (3.23) for the electric and magnetic fields. The geometric view derives it as follows: (1) Orient the axes of the two frames so their relative motion is in the

z

-direction. (2) Perform a simple Lorentz transformation (equation 2.45) on the components of the electromagnetic field tensor:

\begin{aligned} {\bar{E}}_{‖} = {\bar{E}}_{z} = F_{\overset{―}{30}} = Λ^{α}_{3} Λ^{β}_{\overset{―}{0}} F_{α β} = γ^{2} F_{30} + β^{2} γ^{2} F_{03} \\ (3.24) & = (1 - β^{2}) γ^{2} F_{30} = F_{30} = E_{z} = E_{‖}, \\ {\bar{E}}_{x} = F_{\overset{―}{10}} = Λ^{α}_{1} Λ^{β}_{\overset{―}{0}} F_{α β} = γ F_{10} + β γ F_{13} = γ (E_{x} - β B_{y}), \\ etc. \end{aligned}

By contrast, the

3 + 1

view shows much more work. A standard approach is based on the Lorentz force law and energy-change law (

3.2 a, b

), written in the slightly modified form

\begin{matrix} (3.25) & m \frac{d^{2} \bar{x}}{d τ^{2}} = e ({\bar{E}}_{x} \frac{d \bar{t}}{d τ} + 0 \frac{d \bar{x}}{d τ} + {\bar{B}}_{z} \frac{d \bar{y}}{d τ} - {\bar{B}}_{y} \frac{d \bar{z}}{d τ}) \end{matrix}

... (three additional equations) ....
It proceeds as follows (details omitted because of their great length!):
(1) Substitute for the

d^{2} \bar{x} / d τ^{2}

, etc., the expression for these quantities in terms of the

d^{2} x / d τ^{2}, \dots

(Lorentz transformation).
(2) Substitute for the

d^{2} x / d τ^{2}, \dots

the expression for these accelerations in terms of the laboratory

E

and

B

(Lorentz force law).
(3) In these expressions, wherever the components

d x / d τ

of the 4 -velocity in the laboratory frame appear, substitute expressions in terms of the 4 -velocities in the rocket frame (inverse Lorentz transformation).
(4) In (3.25) as thus transformed, demand equality of left and right sides for all values of the

d \bar{x} / d τ

, etc. (validity for all test particles).
(5) In this way arrive at the expressions (3.23) for the

\overset{―}{E}

and

\overset{―}{B}

in terms of the

E

and

B

The contrast in difficulty is obvious!

§3.4. MAXVELL'S EQUATIONS

Turn now from the action of the field on a charge, and ask about the action of a charge on the field, or, more generally, ask about the dynamics of the electromagnetic

Magnetodynamics derived from magnetostatics

Magnetodynamics and magnetostatics unified in one geometric law
field, charge or no charge. Begin with the simplest of Maxwell's equations in a specific Lorentz frame, the one that says there are no free magnetic poles:

\begin{matrix} (3.26) & \nabla \cdot B \equiv div B = \frac{\partial B_{x}}{\partial x} + \frac{\partial B_{y}}{\partial y} + \frac{\partial B_{z}}{\partial z} = 0 . \end{matrix}

This statement has to be true in all Lorentz frames. It is therefore true in the rocket frame:

\begin{matrix} (3.27) & \frac{\partial {\bar{B}}_{x}}{\partial \bar{x}} + \frac{\partial {\bar{B}}_{y}}{\partial \bar{y}} + \frac{\partial {\bar{B}}_{z}}{\partial \bar{z}} = 0 . \end{matrix}

For an infinitesimal Lorentz transformation in the

x

-direction (nonrelativistic velocity

β

), one has (see Box 2.4 and equations 3.23)

\begin{matrix} (3.28) & {\bar{B}}_{x} = B_{x}, {\bar{B}}_{y} = B_{y} + β E_{z}, {\bar{B}}_{z} = B_{z} - β E_{y} \\ (3.29) & \frac{\partial}{\partial \bar{x}} = \frac{\partial}{\partial x} + β \frac{\partial}{\partial t}, \frac{\partial}{\partial \bar{y}} = \frac{\partial}{\partial y}, \frac{\partial}{\partial \bar{z}} = \frac{\partial}{\partial z} \end{matrix}

Substitute into the condition of zero divergence in the rocket frame. Recover the original condition of zero divergence in the laboratory frame, plus the following additional information (requirement for the vanishing of the coefficient of the arbitrary small velocity

β

\begin{matrix} (3.30) & \frac{\partial B_{x}}{\partial t} + \frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z} = 0 . \end{matrix}

Had the velocity of transformation been directed in the

y

- or

z

-directions, a similar equation would have been obtained for

\partial B_{y} / \partial t

\partial B_{z} / \partial t

. In the language of threedimensional vectors, these three equations reduce to the one equation

\begin{matrix} (3.31) & \frac{\partial B}{\partial t} + \nabla \times E \equiv \frac{\partial B}{\partial t} + curl E = 0 \end{matrix}

How beautiful that (1) the principle of covariance (laws of physics are the same in every Lorentz reference system, which is equivalent to the geometric view of physics) plus (2) the principle that magnetic tubes of force never end, gives (3) Maxwell's dynamic law for the time-rate of change of the magnetic field! This suggests that the magnetostatic law

\nabla \cdot B = 0

and the magnetodynamic law

\partial B / \partial t + \nabla \times E = 0

must be wrapped up together in a single frame-independent, geometric law. In terms of components of the field tensor

F

, that geometric law must read

\begin{matrix} (3.32) & F_{α β, γ} + F_{β γ, α} + F_{γ α, β} = 0, \end{matrix}

since this reduces to

\nabla \cdot B = 0

when one takes

α = 1, β = 2, γ = 3

; and it reduces to

\partial B / \partial t + \nabla \times E = 0

when one sets any index, e.g.,

α

, equal to zero (see exercise 3.7 below). In frame-independent geometric language, this law is written (see

§ 3.5

, exercise 3.14, and Chapter 4 for notation)

\begin{matrix} (3.33) & d F = 0, or, equivalently, \nabla \cdot * F = 0; \end{matrix}

and it says, "Take the electromagnetic 2 -form

F

(a geometric object defined even in absence of coordinates); from it construct a new geometric object

d F

(called the "exterior derivative of

F

");

d F

must vanish. The details of this coordinate-free process will be spelled out in exercise 3.15 and in

§ 4.5

(track 2).

Two of Maxwell's equations remain: the electrostatic equation

\begin{matrix} (3.34) & \nabla \cdot E = 4 π ρ, \end{matrix}

and the electrodynamic equation

\begin{matrix} (3.35) & \partial E / \partial t - \nabla \times B = - 4 π J . \end{matrix}

They, like the magnetostatic and magnetodynamic equations, are actually two different parts of a single geometric law. Written in terms of field components, that law says

\begin{matrix} (3.36) & F_{, β}^{α β} = 4 π J^{α}, \end{matrix}

where the components of the "4-current"

J

are

\begin{aligned} J^{0} = ρ & = charge density \\ (3.37) & (J^{1}, J^{2}, J^{3}) & = components of current density. \end{aligned}

Written in coordinate-free, geometric language, this electrodynamic law says

\begin{matrix} (3.38) & d^{*} F = 4 π^{*} J or, equivalently, \nabla \cdot F = 4 π J \end{matrix}

(For full discussion, see exercise 3.15 and

§ 4.5

, which is on Track 2.)

Exercise 3.7. MAXWELL'S EQUATIONS

EXERCISE

Show, by explicit examination of components, that the geometric laws

F_{α β, γ} + F_{β γ, α} + F_{γ α, β} = 0, F^{α β}_{, β} = 4 π J^{α},

do reduce to Maxwell's equations (3.26), (3.31), (3.34), (3.35), as claimed above.

§3.5 WORKING WITH TENSORS

Another mathematical digression is needed. Given an arbitrary tensor field,

s

, of arbitrary rank (choose rank

= 3

for concreteness), one can construct new tensor fields by a variety of operations.

One operation is the gradient

\nabla

. (The symbol

d

is reserved for gradients of scalars,
Ways to produce new tensors from old:
Gradient
in which case

\nabla f \equiv d f

, and for "exterior derivatives of differential forms;" a Track-2

Electrodynamics and electrostatics unified in one geometric law
concept, on which see §4.5.) Like

S, \nabla S

is a machine. It has four slots, whereas

S

has three. It describes how

S

changes from point to point. Specifically, if one desires to know how the number

S (u, v, w)

for fixed

u, v, w

changes under a displacement

ξ

, one inserts

u, v, w, ξ

into the four slots of

\nabla S

\begin{aligned} \nabla S (u, v, w, ξ) \equiv & \partial_{ξ} S (u, v, w) with u, v, w fixed \\ (3.39) & ≃ & + [value of S (u, v, w) at tip of ξ] \\ - [value of S (u, v, w) at tail of ξ] . \end{aligned}

In component notation in a Lorentz frame, this says

\begin{aligned} \nabla S (u, v, w, ξ) & \equiv \partial_{ξ} (S_{α β γ} u^{α} v^{β} w^{γ}) = (\frac{\partial S_{α β γ}}{\partial x^{δ}} ξ^{δ}) u^{α} v^{β} w^{γ} \\ = S_{α β γ, δ} u^{α} v^{β} w^{γ} ξ^{δ} . \end{aligned}

Thus, the Lorentz-frame components of

\nabla S

are nothing but the partial derivatives of the components of

S

. Notice that the gradient raises the rank of a tensor by 1 (from 3 to 4 for

S

Contraction is another process that produces a new tensor from an old one. It seals off ("contracts") two of the old tensor's slots, thereby reducing the rank by two. Specifically, if

R

is a fourth-rank tensor and

M

is obtained by contracting the first and third slots of

R

, then the output of

M

is given by (definition!)

\begin{matrix} (3.40) & M (u, v) = \sum_{α = 0}^{3} R (e_{α}, u, ω^{α}, v) \end{matrix}

Here

e_{α}

and

ω^{α}

are the basis vectors and 1-forms of a specific but arbitrary Lorentz coordinate frame. It makes no difference which frame is chosen; the result will always be the same (exercise 3.8 below). In terms of components in any Lorentz frame, equation (3.40) says (exercise 3.8)

M (u, v) = M_{μ ν} u^{μ} v^{ν} = R_{α μ}_{α}^{α}_{ν} u^{μ} v^{ν}

so that

\begin{matrix} (3.41) & M_{μ ν} = R_{α μ}^{α}_{p} . \end{matrix}

Thus, in terms of components, contraction amounts to putting one index up and the other down, and then summing on them.
Divergence is a third process for creating new tensors from old. It is accomplished by taking the gradient, then contracting the gradient's slot with one of the original slots:
(divergence of

S

on first slot)

\equiv \nabla \cdot S

is a machine such that

\begin{matrix} (3.42) & \nabla \cdot S (u, v) = \nabla S (ω^{α}, u, v, e_{α}) = S^{α}_{β γ, α} u^{β} v^{γ}; \end{matrix}

i.e.

\nabla \cdot S

has components

S^{α}_{β γ, α}

Transpose is a fourth, rather trivial process for creating new tensors. It merely interchanges two slots:

N

obtained by transposing second and third slots of

S \Rightarrow

\begin{matrix} (3.43) & N (u, v, w) = S (u, w, v) \end{matrix}

Symmetrization and antisymmetrization are fifth and sixth processes for producing new tensors from old. A tensor is completely symmetric if its output is unaffected by an interchange of two input vectors or 1 -forms:

\begin{matrix} (3.44a) & S (u, v, w) = S (v, u, w) = S (v, w, u) = \dots \end{matrix}

It is completely antisymmetric if it reverses sign on each interchange of input

\begin{matrix} (3.44b) & S (u, v, w) = - S (v, u, w) = + S (v, w, u) = \dots \end{matrix}

Any tensor can be symmetrized or antisymmetrized by constructing an appropriate linear combination of it and its transposes; see exercise 3.12 .

Wedge product is a seventh process for producing new tensors from old. It is merely an antisymmetrized tensor product: given two vectors

u

and

v

, their wedge product, the "bivector"

u \land v

, is defined by

\begin{matrix} (3.45a) & u \land v \equiv u \otimes v - v \otimes u \end{matrix}

similarly, the "2-form"

α \land β

constructed from two 1-forms is

\begin{matrix} (3.45b) & α \land β \equiv α \otimes β - β \otimes α \end{matrix}

From three vectors

u, v, w

one constructs the "trivector"

\begin{aligned} u \land v \land w \equiv & (u \land v) \land w \equiv u \land (v \land w) \\ = & u \otimes v \otimes w + terms that guarantee complete antisymmetry \\ (3.45c) & = & u \otimes v \otimes w + v \otimes w \otimes u + w \otimes u \otimes v \\ - v \otimes u \otimes w - u \otimes w \otimes v - w \otimes v \otimes u \end{aligned}

From 1-forms

α, β, γ

one similarly constructs the "3-forms"

α \land β \land γ

. The wedge product gives a simple way to test for coplanarity (linear dependence) of vectors: if

u

and

v

are collinear, so

u = a v

, then

u \land v = a v \land v = 0 (by antisymmetry of " \land ")

w

is coplanar with

u

and

v

w = a u + b v

("collapsed box"), then

w \land u \land v = a u \land u \land v + b v \land u \land v = 0

The symbol "

\land

" is called a "hat" or "wedge" or "exterior product sign." Its properties are investigated in Chapter 4.

Taking the dual is an eighth process for constructing new tensors. It plays a Dual fundamental role in Track 2 of this book, but since it is not needed for Track 1, its definition and properties are treated only in the exercises ( 3.14 and 3.15 ).

Symmetrization and antisymmetrization

Wedge product

Bivector

2-form

Trivector

Because the frame-independent geometric notation is somewhat ambiguous (which slots are being contracted? on which slot is the divergence taken? which slots are being transposed?), one often uses component notation to express coordinate-independent, geometric relations between geometric objects. For example,

J_{β γ} = S_{β γ, α}^{α}

means "

J

is a tensor obtained by taking the divergence on the first slot of the tensor

S^{''}

. Also,

v^{γ} = {(F_{μ ν} F^{μ ν})}^{, γ} \equiv {(F_{μ ν} F^{μ ν})}_{, β} η^{β γ}

means "

v

is a vector obtained by (1) constructing the tensor product

F \otimes F

F

with itself, (2) contracting

F \otimes F

on its first and third slots, and also on its second and fourth, (3) taking the gradient of the resultant scalar function, (4) converting that gradient, which is a 1 -form, into the corresponding vector."
Index gymnastics
"Index gymnastics," the technique of extracting the content from geometric equations by working in component notation and rearranging indices as required, must be mastered if one wishes to do difficult calculations in relativity, special or general. Box 3.3 expounds some of the short cuts in index gymnastics, and exercises 3.8-3.18 offer practice.

EXERCISES

Exercise 3.8. CONTRACTION IS FRAME-INDEPENDENT

Show that contraction, as defined in equation (3.40), does not depend on which Lorentz frame

e_{α}

and

ω^{α}

are taken from. Also show that equation (3.40) implies

M (u, v) = R_{α μ}^{α}_{ν} u^{μ} v^{v}

Exercise 3.9. DIFFERENTIATION

(a) Justify the formula

d (u_{μ} v^{v}) / d τ = (d u_{μ} / d τ) v^{v} + u_{μ} (d v^{v} / d τ)

by considering the special case

μ = 0, ν = 1

.
(b) Explain why

{(T^{α β} v_{β})}_{, μ} = T_{, μ}^{α β} v_{β} + T^{α β} v_{β, μ}

Exercise 3.10. MORE DIFFERENTIATION

(a) Justify the formula,

d (u^{μ} u_{μ}) / d τ = 2 u_{μ} (d u^{μ} / d τ)

by writing out the summation

u^{μ} u_{μ} \equiv η_{μ ν} u^{μ} u^{ν}

explicitly.
(b) Let

δ

indicate a variation or small change, and justify the formula

δ (F_{α β} F^{α β}) = 2 F_{α β} δ F^{α β} .

{(F_{α β} F^{α β})}_{, μ} =

Box 3.3 TECHNIQUES OF INDEX GYMNASTICS

Equation

Name and Discussion

\begin{aligned} S^{α}_{β γ} = S (ω^{α}, e_{β}, e_{γ}) \\ S^{α β}_{γ} = S (ω^{α}, ω^{β}, e_{γ}) \\ S = S^{α}_{β γ} e_{α} \otimes ω^{β} \otimes ω^{γ} \\ S = S^{α β γ} e_{α} \otimes e_{β} \otimes e_{γ} \\ S^{α β}_{γ} = η^{β μ} S^{α}_{μ γ} \\ S^{α}_{μ γ} = η_{μ β} S^{α β}_{γ} \\ M_{μ} = S^{α}_{μ α} \\ T^{α β}_{μ ν} = S^{α β}_{μ} M_{v} \\ A^{2} = A^{α} A_{α} \\ η_{α β} η^{β γ} = δ_{α}^{γ} \\ S^{α}_{β γ} = N^{α}_{β, γ} \\ R_{β} = N^{α}_{β, α} \\ N^{α}_{β, γ} = {(η_{β μ} N^{α μ})}_{, γ} = η_{β μ} N^{α μ}_{, γ} \\ N^{α}_{β}^{γ} \equiv N^{α}_{β, μ} η^{μ γ} \\ {(R_{α} M_{β})}_{, γ} = R_{α, γ} M_{β} + R_{α} M_{β, γ} \\ G_{α β} = F_{[α β]} \equiv \frac{1}{2} (F_{α β} - F_{β α}) \\ H_{α β} = F_{(α β)} \equiv \frac{1}{2} (F_{α β} + F_{β α}) \\ ^{*} J_{α β γ} = J^{μ} ε_{μ α β γ} \\ ^{*} F_{α β} = \frac{1}{2} F^{μ v} ε_{μ ν α β} \\ ^{*} B_{α} =_{6}^{1} B^{λ μ ν} ε_{λ μ ν α} \end{aligned}

Computing other components.
Reconstructing the rank-(

\frac{1}{2}

) version of

S

.
Reconstructing the rank-

(\binom{3}{0})

version of

S

. [Recall: one does not usually distinguish between the various versions; see equation (3.15) and associated discussion.]

Raising an index.
Lowering an index.
Contraction of

S

to form a new tensor

M

.
Tensor product of

S

with

M

to form a new tensor

T

.
Squared length of vector

A

produced by forming tensor product

A \otimes A

and then contracting, which is the same as forming the corresponding 1-form

\tilde{A}

and then piercing:

A^{2} = ⟨ \tilde{A}, A ⟩ = A^{α} A_{α}

The matrix formed from the metric's "covariant components,"

‖ η_{α β} ‖

, is the inverse of that formed from its "contravariant components,"

‖ η^{α β} ‖

. Equivalently, raising one index of the metric

η_{α β}

produces the Kronecker delta.

Gradient of

N

to form a new tensor

S

.
Divergence of

N

to form a new tensor

R

.
Taking gradients and raising or lowering indices are operations that commute.

Contravariant index on a gradient is obtained by raising covariant index.

Gradient of a tensor product; says

\nabla (R \otimes M) =

Transpose

(\nabla R \otimes M) + R \otimes \nabla M

.
Antisymmetrizing a tensor

F

to produce a new tensor

G

.
Symmetrizing a tensor

F

to produce a new tensor

H

.
Forming the rank-3 tensor that is dual to a vector (exercise 3.14).

Forming the antisymmetric rank-2 tensor that is dual to a given antisymmetric rank-2 tensor (exercise 3.14).

Forming the 1 -form that is dual to an antisymmetric rank-3 tensor (exercise 3.14).

Exercise 3.11. SYMMETRIES

Let

A_{μ ν}

be an antisymmetric tensor so that

A_{μ ν} = - A_{ν μ}

; and let

S^{μ ν}

be a symmetric tensor so that

S^{μ ν} = S^{ν μ}

.
(a) Justify the equations

A_{μ ν} S^{μ ν} = 0

in two ways: first, by writing out the sum explicitly (all sixteen terms) and showing how the terms in the sum cancel in pairs; second, by giving an argument to justify each equals sign in the following string:

A_{μ ν} S^{μ ν} = - A_{ν μ} S^{μ ν} = - A_{ν μ} S^{ν μ} = - A_{α β} S^{α β} = - A_{μ ν} S^{μ ν} = 0

(b) Establish the following two identities for any arbitrary tensor

V_{μ ν}

V^{μ ν} A_{μ ν} = \frac{1}{2} (V^{μ ν} - V^{ν μ}) A_{μ ν}, V^{μ ν} S_{μ ν} = \frac{1}{2} (V^{μ ν} + V^{ν μ}) S_{μ ν} .

Exercise 3.12. SYMMETRIZATION AND ANTISYMMETRIZATION

To "symmetrize" a tensor, one averages it with all of its transposes. The components of the new, symmetrized tensor are distinguished by round brackets:

\begin{aligned} V_{(μ ν)} & \equiv \frac{1}{2} (V_{μ ν} + V_{ν μ}) \\ (3.46) & V_{(μ ν λ)} & \equiv \frac{1}{3!} (V_{μ ν λ} + V_{ν λ μ} + V_{λ μ ν} + V_{ν μ λ} + V_{μ λ ν} + V_{λ ν μ}) \end{aligned}

One "antisymmetrizes" a tensor (square brackets) similarly:

\begin{aligned} V_{[μ ν]} & \equiv \frac{1}{2} (V_{μ ν} - V_{ν μ}) \\ (3.47) & V_{[μ ν λ]} & \equiv \frac{1}{3!} (V_{μ ν λ} + V_{ν λ μ} + V_{λ μ ν} - V_{ν μ λ} - V_{μ λ ν} - V_{λ ν ν}) . \end{aligned}

(a) Show that such symmetrized and antisymmetrized tensors are, indeed, symmetric and antisymmetric under interchange of the vectors inserted into their slots:

\begin{aligned} V_{(α β γ)} u^{α} v^{β} w^{γ} = + V_{(α β γ} v^{α} u^{β} w^{γ} = \dots, \\ V_{[α β γ]} u^{α} v^{β} w^{γ} = - V_{[α β γ]} v^{α} u^{β} w^{γ} = \dots \end{aligned}

(b) Show that a second-rank tensor can be reconstructed from its symmetric and antisymmetric parts,

\begin{matrix} (3.48) & V_{μ ν} = V_{(μ ν)} + V_{[μ ν]}, \end{matrix}

but that a third-rank tensor cannot;

V_{(α β γ)}

and

V_{[α β γ]}

contain together "less information" than

V_{α β γ}

"Young diagrams" (see, e.g., Messiah [1961], appendix D) describe other symmetries, more subtle than these two, which contain the missing information.
(c) Show that the electromagnetic field tensor satisfies

\begin{matrix} (3.49a) & F_{(α β)} = 0, F_{α β} = F_{[α β]} . \end{matrix}

(d) Show that Maxwell's "magnetic" equations

F_{α β, γ} + F_{β γ, α} + F_{γ α, β} = 0

can be rewritten in the form

\begin{matrix} (3.49b) & F_{[α β, γ]} = 0 . \end{matrix}

Exercise 3.13. LEVI-CIVITA TENSOR

The "Levi-Civita tensor"

ε

in spacetime is a fourth-rank, completely antisymmetric tensor:

ε (n, u, v, w)

changes sign when any two of the
vectors are interchanged.

Choose an arbitrary but specific Lorentz frame, with

e_{0}

pointing toward the future and with

e_{1}, e_{2}, e_{3}

a righthanded set of spatial basis vectors. The covariant components of

ε

in this frame are

\begin{matrix} (3.50b) & ε_{0123} = ε (e_{0}, e_{1}, e_{2}, e_{3}) = + 1 \end{matrix}

[Note: In an

n

-dimensional space,

ε

is the analogous completely antisymmetric rank-

n

tensor. Its components are

\begin{matrix} (3.50c) & ε_{12 \dots n} = ε (e_{1}, e_{2}, \dots, e_{n}) = + 1 \end{matrix}

when computed on a "positively oriented," orthonormal basis

e_{1}, \dots, e_{n}

.]
(a) Use the antisymmetry to show that

\begin{aligned} (3.50~d) & ε_{α β γ δ} & = 0 unless α, β, γ, δ are all different, \\ (3.50e) & ε_{π 0 π_{1} π_{2} π_{3}} & = {\begin{matrix} + 1 for even permutations of 0, 1, 2, 3, and \\ - 1 for odd permutations. \end{matrix} \end{aligned}

(b) Show that

\begin{matrix} (3.50f) & ε^{π_{0} π_{1} π_{2} π_{3}} = - ε_{π_{0} π_{1} π_{2} π_{3}} . \end{matrix}

ε^{\bar{α} \bar{β} γ \bar{δ}}

and

ε_{\bar{α} \bar{β} \bar{γ} \bar{δ}}

have these same values in any other Lorentz frame with

e_{\overset{―}{0}}

pointing toward the future and with

e_{\overset{―}{1}}, e_{\overset{―}{2}}, e_{\overset{―}{3}}

a righthanded set. Hint: show that

\begin{matrix} (3.50~g) & ε^{α β γ δ} Λ^{\overset{―}{0}}_{α} Λ^{\overset{―}{1}} Λ^{\overset{―}{2}}_{γ} Λ^{\overset{―}{3}}_{δ} = - \det | Λ^{\bar{μ}}_{ν} |; \end{matrix}

from

Λ^{T} η Λ = η

, show that

\det | Λ^{\bar{μ}}_{ν} | = \pm 1

; and verify that the determinant is +1 for transformations between frames with

e_{0}

and

e_{0}

future-pointing, and with

e_{1}, e_{2}, e_{3}

and

e_{\overset{―}{1}}, e_{\overset{―}{2}}

e_{\overset{―}{3}}

righthanded.
(d) What are the components of

ε

in a Lorentz frame with past-pointing

e_{\overset{―}{0}}

? with lefthanded

e_{\overset{―}{1}}, e_{\overset{―}{2}}, e_{\overset{―}{3}}

?
(e) From the Levi-Civita tensor, one can construct several "permutation tensors." In index notation:

\begin{aligned} (3.50h) & δ^{α β γ}_{μ ν λ} & \equiv - ε^{α β γ ρ} ε_{μ ν λ ρ} \\ (3.50i) & δ^{α β}_{μ ν} & \equiv \frac{1}{2} δ^{α β λ}_{μ ν λ} = - \frac{1}{2} ε^{α β λ ρ} ε_{μ ν λ ρ} \\ (3.50j) & δ^{α}_{μ} & \equiv \frac{1}{3} δ^{α β}_{μ β} = \frac{1}{6} δ^{α β λ}_{μ β λ} = - \frac{1}{6} ε^{α β λ ρ} ε_{μ β λ ρ} . \end{aligned}

Show that:

δ^{α β γ}_{μ ν λ} = {\begin{aligned} (3.50k) & + 1 & if α β γ is an even permutation of μ ν λ, \\ - 1 & if α β γ is an odd permutation of μ ν λ, \\ 0 & otherwise; \end{aligned}

\begin{aligned} δ^{α β}_{μ ν} = δ^{α}_{μ}^{β}^{β}_{ν} - δ^{α}_{ν}^{δ}^{β}_{μ} \\ (3.501) & = {\begin{matrix} + 1 if α β is an even permutation of μ ν, \\ - 1 if α β is an odd permutation of μ ν, \\ 0 otherwise; \end{matrix} \\ (3.50~m) & δ^{α}_{μ} = {\begin{array}{r} + 1 if α = μ, \\ 0 otherwise. \end{array} \end{aligned}

Exercise 3.14. DUALS

From any vector

J

, any second-rank antisymmetric tensor

F (F_{α β} = F_{[α β]})

, and any third-rank antisymmetric tensor

B (B_{α β γ} = B_{[α β γ]})

, one can construct new tensors defined by

\begin{matrix} (3.51) & ^{*} J_{α β γ} = J^{μ} ε_{μ α β γ},^{*} F_{α β} = \frac{1}{2} F^{μ ν} v_{μ ν α β},^{*} B_{α} = \frac{1}{3!} B^{λ μ ν} ε_{λ μ ν α} . \end{matrix}

One calls

J

the "dual" of

J,^{*} F

the dual of

F

, and

* B

the dual of

B

. [A previous and entirely distinct use of the word "dual" (

$ 2.7

) called a set of basis one-forms

{ω^{α}}

dual to a set of basis vectors

{e_{α}}

⟨ ω^{α}, e_{β} ⟩ = δ^{α}_{β}

. Fortunately there are no grounds for confusion between the two types of duality. One relates sets of vectors to sets of 1 -forms. The other relates antisymmetric tensors of rank

p

to antisymmetric tensors of rank

4 - p

.]
(a) Show that

\begin{matrix} (3.52) & ^{* *} J = J,^{* *} F = - F,^{* *} B = B \end{matrix}

so (aside from sign) one can recover any completely antisymmetric tensor

H

from its dual

^{*} H

by taking the dual once again,

^{* *} H

. This shows that

H

and

^{* H}

contain precisely the same information.
(b) Make explicit this fact of same-information-content by writing out the components

^{*} A^{α β γ}

in terms of

A^{α}

, also

^{*} F^{α β}

in terms of

F^{α β}

, also

^{*} B^{α}

in terms of

B^{α β γ}

Exercise 3.15. GEOMETRIC VERSIONS OF MAXWELL EQUATIONS

Show that, if

F

is the electromagnetic field tensor, then

\nabla \cdot^{*} F = 0

is a geometric frame-independent version of the Maxwell equations

F_{α β, γ} + F_{β γ, α} + F_{γ α, β} = 0

Similarly show that

\nabla \cdot F = 4 π J

(divergence on second slot of

F

) is a geometric version of

F^{α β}_{, β} = 4 π J^{α}

Exercise 3.16. CHARGE CONSERVATION

From Maxwell's equations

F^{α β}_{, β} = 4 π J^{α}

, derive the "equation of charge conservation"

\begin{matrix} (3.53) & J^{α}_{, α} = 0 . \end{matrix}

Show that this equation does, indeed, correspond to conservation of charge. It will be studied further in Chapter 5.

Exercise 3.17. VECTOR POTENTIAL

The vector potential

A

of electromagnetic theory generates the electromagnetic field tensor via the geometric equation

\begin{matrix} (3.54) & F = - (antisymmetric part of \nabla A), \end{matrix}

i.e.,

\begin{matrix} (') & F_{μ ν} = A_{ν, μ} - A_{μ, ν} . \end{matrix}

(a) Show that the electric and magnetic fields in a specific Lorentz frame are given by

\begin{matrix} (3.55) & B = \nabla \times A, E = - \partial A / \partial t - \nabla A^{0} \end{matrix}

(b) Show that

F

will satisfy Maxwell's equations if and only if

A

satisfies

\begin{matrix} (3.56) & A_{, μ}^{α, μ} - A_{, μ}^{μ, α} = - 4 π J^{α} \end{matrix}

\begin{matrix} (3.57) & A_{NEW} = A_{OLD} + d ϕ, ϕ = arbitrary function \end{matrix}

leave

F

unaffected.
(d) Show that one can adjust the gauge so that

\begin{aligned} (3.58a) & \nabla \cdot A & = 0 ("Lorentz gauge") \\ (3.58b) & ◻ A & = - 4 π J . \end{aligned}

Hereis the wave operator ("d'Alembertian"):

\begin{matrix} (3.59) & ◻ A = A_{, μ}^{α, μ} e_{α} \end{matrix}

Exercise 3.18. DIVERGENCE OF ELECTROMAGNETIC STRESS-ENERGY TENSOR

From an electromagnetic field tensor

F

, one constructs a second-rank, symmetric tensor

T

("stress-energy tensor," to be studied in Chapter 5) as follows:

\begin{matrix} (3.60) & T^{μ ν} = \frac{1}{4 π} (F^{μ α} F_{α}^{ν} - \frac{1}{4} η^{μ ν} F_{α β} F^{α β}) \end{matrix}

As an exercise in index gymnastics:
(a) Show that

\nabla \cdot T

has components

\begin{matrix} (3.61) & T^{μ ν}_{, ν} = \frac{1}{4 π} [F^{μ α}_{, ν} F^{v}_{α} + F^{μ α} F_{α, ν}^{ν} - \frac{1}{2} F_{α β}, μ F^{α β}] . \end{matrix}

(b) Manipulate this expression into the form

\begin{matrix} (3.62) & T_{μ}^{ν}, ν = \frac{1}{4 π} [- F_{μ α} F^{α ν}_{, ν} - \frac{1}{2} F^{α β} (F_{α β, μ} + F_{μ α, β} + F_{β μ, α})] \end{matrix}

note that the first term of (3.62) arises directly from the second term of (3.61).
(c) Use Maxwell's equations to conclude that

\begin{matrix} (3.63) & T^{μ ν}_{, p} = - F^{μ α} J_{α} \end{matrix}

ELECTROMAGNETISM AND DIFFERENTIAL FORMS

The ether trembled at his agitations In a manner so familiar that I only need to say, In accordance with Clerk Maxwell's six equations

It tickled peoples' optics far away. You can feel the way it's done, You may trace them as they run-

d γ

by dy less

d β

by dz is equal

KdX / dt

While the curl of

(X, Y, Z)

is the minus

d / dt

of the vector

(a, b, c)

From The Revolution of the Corpuscle,
written by A. A. Robb
(to the tune of The Interfering Parrott) for a dinner of the research students of the Cavendish Laboratory in the days of the old mathematics.

This chapter is all Track 2. It is needed as preparation for

§ § 14.5

and 14.6 (computation of curvature using differential forms) and for Chapter 15 (Bianchi identities and boundary of a boundary), but is not needed for the rest of the book.

§4.1. EXTERIOR CALCULUS

Stacks of surfaces, individually or intersecting to make "honeycombs," "egg crates," and other such structures ("differential forms"), give unique insight into the geometry of electromagnetism and gravitation. However, such insight comes at some cost in time. Therefore, most readers should skip this chapter and later material that depends on it during a first reading of this book.

Analytically speaking, differential forms are completely antisymmetric tensors; pictorially speaking, they are intersecting stacks of surfaces. The mathematical formalism for manipulating differential forms with ease, called "exterior calculus," is summarized concisely in Box 4.1 ; its basic features are illustrated in the rest of this chapter by rewriting electromagnetic theory in its language. An effective way to tackle this chapter might be to (1) scan Box 4.1 to get the flavor of the formalism; (2) read the rest of the chapter in detail; (3) restudy Box 4.1 carefully; (4) get practice in manipulating the formalism by working the exercises.*
(continued on page 99)
*Exterior calculus is treated in greater detail than here by: E. Cartan (1945); de Rham (1955); Nickerson, Spencer, and Steenrod (1959); Hauser (1970); Israel (1970); especially Flanders (1963, relatively easy, with many applications); Spivak (1965, sophomore or junior level, but fully in tune with modern mathematics); H. Cartan (1970); and Choquet-Bruhat (1968a).

Box 4.1 DIFFERENTIAL FORMS AND EXTERIOR CALCULUS IN BRIEF

The fundamental definitions and formulas of exterior calculus are summarized here for ready reference. Each item consists of a general statement (at left of page) plus a leading application (at right of page). This formalism is applicable not only to spacetime, but also to more general geometrical systems (see heading of each section). No attempt is made here to demonstrate the internal consistency of the formalism, nor to derive it from any set of definitions and axioms. For a systematic treatment that does so, see, e.g., Spivak (1965), or Misner and Wheeler (1957).

A. Algebra I (applicable to any vector space)

Basis 1-forms.
a. Coordinate basis $ω^{j} = d x^{j}$
( $j$ tells which 1-form, not which component).
b. General basis $ω^{j} = L_{k^{'}}^{j} d x^{k^{'}}$ .

An application

Simple basis 1 -forms for analyzing Schwarzschild geometry around static spherically symmetric center of attraction:

\begin{aligned} ω^{0} = (1 - 2 m / r)^{1 / 2} d t; \\ ω^{1} = (1 - 2 m / r)^{- 1 / 2} d r; \\ ω^{2} = r d θ; \\ ω^{3} = r \sin θ d ϕ . \end{aligned}

General p-form (or p-vector) is a completely antisymmetric tensor of rank $(\binom{0}{p}) [$ or $(\binom{p}{0})]$ . It can be expanded in terms of wedge products (see $§ 3.5$ and exercise 4.12):

\begin{aligned} α & = \frac{1}{p!} α_{i_{1} i_{2} \dots i_{p}} ω^{i_{1}} \land ω^{i_{2}} \land \dots \land ω^{i_{p}} \\ \equiv α_{∣ i_{1} i_{2} \dots i_{p}} ω^{i_{1}} \land ω^{i_{2}} \land \dots \land ω^{i_{p}} . \end{aligned}

(Note: Vertical bars around the indices mean summation extends only over

i_{1} < i_{2} < \dots < i_{p}

Two applications

Energy-momentum 1-form is of type

α = α_{i} ω^{i}

p = - E d t + p_{x} d x + p_{y} d y + p_{z} d z

Faraday is a 2 -form of type

β = β_{| μ ν |} ω^{μ} \land ω^{ν}

or in flat spacetime

\begin{aligned} F = & - E_{x} d t \land d x - E_{y} d t \land d y - E_{z} d t \land d z \\ + B_{x} d y \land d z + B_{y} d z \land d x + B_{z} d x \land d y \end{aligned}

Box 4.1 (continued)
3. Wedge product.

All familiar rules of addition and multiplication hold, such as

\begin{aligned} (a α + b β) \land γ = a α \land γ + b β \land γ, \\ (α \land β) \land γ = α \land (β \land γ) \equiv α \land β \land γ, \end{aligned}

except for a modified commutation law between
a

p

-form

α

and a

q

-form

β

\underset{p}{α} \land \underset{q}{β} = (- 1)^{p q} \underset{q}{β} \land \underset{p}{α} .

Applications to 1-forms

α, β

\begin{aligned} α \land β & = - β \land a, α \land α = 0 \\ α \land β & = (α_{j} ω^{j}) \land (β_{k} ω^{k}) = α_{j} β_{k} ω^{j} \land ω^{k} \\ = \frac{1}{2} (α_{j} β_{k} - β_{j} α_{k}) ω^{j} \land ω^{k} . \end{aligned}

Contraction of p -form on p -vector.
$⟨ \underset{p}{α}, \underset{p}{A} ⟩$

\begin{aligned} = α_{| i_{1} \dots i_{p} |} A^{| j_{1} \dots j_{p} |} ⟨ \underset{[ω^{i_{1}} \land \dots \land ω^{i_{p}}, e_{j_{1}} \land \dots \land e_{j_{p}} ⟩}{\underset{⏟}{\equiv δ_{j_{1} \dots i_{p}}^{i_{p}, i_{p}} (see exercises 3.13 and 4.12)]}} \\ = {α_{| i_{1} \dots i_{p} |} |}^{i_{1} \dots i_{p} .} \end{aligned}

Four applications
a. Contraction of a particle's energy-momentum 1-form

p = p_{α} ω^{α}

with 4-velocity

u = u^{α} e_{α}

of observer (a 1-vector):

- ⟨ p, u ⟩ = - p_{α} u^{α} = energy of particle.

b. Contraction of Faraday 2 -form

F

with bivector

δ P \land Δ P

[where

δ P = (d P / d λ_{1}) Δ λ_{1}

and

Δ P =

(d P / d X_{2}) Δ λ_{2}

are two infinitesimal vectors in a 2 -surface

P (λ_{1}, λ_{2})

, and the bivector represents the surface element they span] is the magnetic flux

Φ = ⟨ F, δ P

\land Δ P ⟩

through that surface element.
c. More generally, a

p

-dimensional parallelepiped with vectors

a_{1}, a_{2}, \dots, a_{p}

for legs has an oriented volume described by the "simple"

p

-vector

a_{1} \land a_{2} \land \dots a_{p}

(oriented because interchange of two legs changes its sign). An egg-crate type of structure with walls made from the hyperplanes of

p

different 1 -forms

σ^{1}

σ^{2}, \dots, σ^{p}

is described by the "simple"

p

-form

σ^{1}

\land σ^{2} \land \dots \land σ^{p}

. The number of cells of

σ^{1} \land

σ^{2} \land \dots \land σ^{p}

sliced through by the infinitesimal

p

-volume

a_{1} \land a_{2} \land \dots \land a_{p}

⟨ σ^{1} \land σ^{2} \land \dots \land σ^{p}, a_{1} \land a_{2} \land \dots \land a_{p} ⟩

d. The Jacobian determinant of a set of

p

functions

f^{k} (x^{1}, \dots, x^{n})

with respect to

p

of their arguments is

\begin{aligned} ⟨ d f^{1} \land d f^{2} \land \dots & \land d f^{p}, \frac{\partial P}{\partial x^{1}} \land \frac{\partial P}{\partial x^{2}} \land \dots \land \frac{\partial P}{\partial x^{p}} ⟩ \\ = \det ‖ (\frac{\partial f^{k}}{\partial x^{j}}) ‖ \equiv \frac{\partial (f^{1}, f^{2}, \dots, f^{p})}{\partial (x^{1}, x^{2}, \dots, x^{p})} . \end{aligned}

Simple forms.
a. A simple $p$ -form is one that can be written as a wedge product of $p 1$ -forms:

\underset{p}{σ} = \underset{p factors}{\underset{⏟}{α \land β \land \dots \land γ}}

b. A simple

p

-form

α \land β \land \dots \land γ

is represented by the intersecting families of surfaces of

α, β, \dots, γ

(egg-crate structure) plus a sense of circulation (orientation).

Applications:

a. In four dimensions (e.g., spacetime) all 0 -forms, 1 forms, 3 -forms, and 4 -forms are simple. A 2 -form

F

is generally a sum of two simple forms, e.g.,

F =

- e d t \land d x + h d y \land d z

; it is simple if and only if

F \land F = 0

.
b. A set of 1 -forms

α, β, \dots, γ

is linearly dependent (one a linear combination of the others) if and only if

α \land β \land \dots \land γ = 0

(egg crate collapsed).
B. Exterior Derivative (applicable to any "differentiable manifold," with or without metric)

$d$ produces a $(p + 1)$ -form $d σ$ from a $p$ -form $σ$ .
Effect of $d$ is defined by induction using the

Box 4.1 (continued)

(Chapter 2) definition of

d f

, and

f

a function (

0 -

form), plus

\begin{aligned} d (\underset{p}{α} \land {\underset{q}{β}}^{β} & = d α \land β + (- 1)^{p} α \land d β \\ d^{2} & = d d = 0 \end{aligned}

Two applications

d (α \land d β) = d α \land d β

For the

p

-form

ϕ

, with

ϕ = ϕ_{| i_{1} \dots i_{i} |} d x^{i_{1}} \land \dots \land d x^{i_{p}}

one has (alternative and equivalent definition of

d ϕ

)

d ϕ = d_{ϕ_{| i_{1} \dots i_{p} |}} \land d x^{i_{1}} \land \dots \land d x^{i_{p}}

C. Integration (applicable to any "differentiable manifold," with or without metric)

Pictorial interpretation.

Text and pictures of Chapter 4 interpret

\int α

(integral of specified 1-form

α

along specified curve from specified starting point to specified end point) as "number of

α

-surfaces pierced on that route"; similarly, they interpret

\int ϕ

(integral of specified 2 -form

ϕ

over specified bit of surface on which there is an assigned sense of circulation or "orientation") as "number of cells of the honeycomb-like structure

ϕ

cut through by that surface"; similarly for the egg-crate-like structures that represent 3forms; etc.
2. Computational rules for integration.

To evaluate

\int α

, the integral of a

p

-form

α = α_{| i_{1} \dots i_{p} |} (x^{1}, \dots, x^{n}) d x^{i_{1}} \land \dots \land d x^{i_{p}}

over a

p

-dimensional surface, proceed in two steps.
a. Substitute a parameterization of the surface,

x^{k} (λ^{1}, \dots, λ^{p})

into

α

, and collect terms in the form

a = a (λ^{j}) d λ^{1} \land \dots \land d λ^{p}

(this is

α

viewed as a

p

-form in the

p

-dimensional surface);
b. Integrate

\int a = \int a (λ^{j}) d λ^{1} d λ^{2} \dots d λ^{p}

using elementary definition of integration.
Example: See equations (4.12) to (4.14).
3. The differential geometry of integration.

Calculate

\int α

for a

p

-form

α

as follows.
a. Choose the

p

-dimensional surface

S

over which to integrate.
b. Represent

S

by a parametrization giving the generic point of the surface as a function of the parameters,

P (λ^{1}, λ^{2}, \dots λ^{p})

. This fixes the orientation. The same function with

λ^{1} \leftrightarrow λ^{2}

P (λ^{2}, λ^{1}, \dots, λ^{p})

, describes a different (i.e., oppositely oriented) surface,

- S

.
c. The infinitesimal parallelepiped

(\frac{\partial P}{\partial λ^{1}} Δ λ^{1}) \land (\frac{\partial P}{\partial λ^{2}} Δ λ^{2}) \land \dots \land (\frac{\partial P}{\partial λ^{p}} Δ λ^{p})

is tangent to the surface. The number of cells of

a

it slices is

⟨ α, \frac{\partial P}{\partial λ^{1}} \land \dots \land \frac{\partial P}{\partial λ^{p}} ⟩ Δ λ^{1} \dots Δ λ^{p}

This number changes sign if two of the vectors

\partial P / \partial λ^{k}

are interchanged, as for an oppositely oriented surface.
d. The above provides an interpretation motivating the definition

\begin{array}{r} \int α \equiv \iint \dots \int ⟨ α, \frac{\partial P}{\partial λ^{1}} \land \frac{\partial P}{\partial λ^{2}} \land \dots \land \frac{\partial P}{\partial λ^{p}} ⟩ \\ d λ^{1} d λ^{2} \dots d λ^{p} \end{array}

This definition is identified with the computational rule of the preceding section (C.2) in exercise 4.9.

An application

Integrate a gradient

d f

along a curve,

P (λ)

from

P (0)

P (1)

\begin{aligned} \int d f & = \int_{0}^{1} ⟨ d f, d P / d λ ⟩ d λ = \int_{0}^{1} (d f / d λ) d λ \\ = f [P (1)] - f [P (0)] \end{aligned}

e. Three different uses for symbol "d": First, lightface

d

in explicit derivative expressions such as

Box 4.1 (continued)

d / d a

, or

d f / d a

, or

d P / d a

; neither numerator nor denominator alone has any meaning, but only the full string of symbols. Second, lightface

d

inside an integral sign; e.g.,

\int f d a

. This is an instruction to perform integration, and has no meaning whatsoever without an integral sign;

" \int \dots d \dots

. lives as an indivisible unit. Third, sans-serif

d

; e.g.,

d

alone, or

d f

, or

d α

. This is an exterior derivative, which converts a

p

-form into a

(p + 1)

-form. Sometimes lightface

d

is used for the same purpose. Hence,

d

alone, or

d f

, or

d x

, is always an exterior derivative unless coupled to an

\int sign

(second use), or coupled to a / sign (first use).
4. The generalized Stokes theorem (see Box 4.6).
a. Let

\partial V

be the closed

p

-dimensional boundary of a

(p + 1)

-dimensional surface

V

. Let

σ

be a

p

-form defined throughout

V

.
Then

\int_{T} d σ = \int_{\partial T} σ

[integral of

p

-form

σ

over boundary

\partial T

equals integral of

(p + 1)

-form

d σ

over interior

V

].
b. For the sign to come out right, orientations of

V

and

\partial V

must agree in this sense: choose coordinates

y^{0}, y^{1}, \dots, y^{p}

on a portion of

T

, with

y^{0}

specialized so

y^{0} \leq 0

V

, and

y^{0} = 0

at the boundary

\partial V

; then the orientation

\frac{\partial P}{\partial y^{0}} \land \frac{\partial P}{\partial y^{1}} \land \dots \land \frac{\partial P}{\partial y^{p}}

for

V

demands the orientation

\frac{\partial P}{\partial y^{1}} \land \dots \land \frac{\partial P}{\partial y^{p}}

for

\partial V

.
c. Note: For a nonorientable surface, such as a Möbius strip, where a consistent and continuous choice of orientation is impossible, more intricate mathematics is required to give a definition of "

\partial

" for which the Stokes theorem holds.

Applications: Includes as special cases all integral theorems for surfaces of arbitrary dimension in spaces of arbitrary dimension, with or without metric, generaliz-
ing all versions of theorems of Stokes and Gauss. Exampes:
a.

V

a curve,

\partial V

its endpoints,

σ = f

a 0 -form (funcion):

\int_{\lor} d f = \int_{0}^{1} (d f / d λ) d λ = \int_{\partial V} f = f (1) - f (0) .

V

a 2 -surface in 3 -space,

\partial V

its closed-curve boundary, va 1 -form; translated into Euclidean vector notation, the two integrals are

\int_{V V} d v = \int_{V V} (\nabla \times v) \cdot d S; \int_{\partial V} v = \int_{\partial γ} v \cdot d I .

c. Other applications in

§ § 5.8, 20.2, 20.3, 20.5

, and exercises 4.10, 4.11, 5.2, and below.

D. Algebra II (applicable to any vector space with metric)

Norm of a p -form.

‖ α ‖^{2} \equiv α_{| i_{1} \dots i_{p} |} α^{α_{1} \dots i_{p}}

Two applications: Norm of a 1-form equals its squared length,

‖ α ‖^{2} = α \cdot α

. Norm of electromagnetic 2-form or Faraday:

‖ F ‖^{2} = B^{2} - E^{2}

.
2. Dual of a p -form.
a. In an

n

-dimensional space, the dual of a

p

-form

α

is the

(n - p)

-form

^{*} α

, with components

{(^{*} α)}_{k_{1} \dots k_{n - p}} = α^{| i_{1} \dots i_{p} |} ϵ_{i_{1} \dots i_{p}} k_{1 \dots k_{n - p}} .

b. Properties of duals:

\begin{aligned} ^{* *} α & = (- 1)^{p - 1} α in spacetime; \\ α \land^{*} α & = ‖ α ‖^{2} ε in general. \end{aligned}

c. Note: the definition of

ε

(exercise 3.13) entails choosing an orientation of the space, i.e., deciding which orthonormal bases (1) are "righthanded" and thus (2) have

ε (e_{1}, \dots, e_{n}) = + 1

Applications

a. For

f

a 0 -form,

^{*} f = f ε

, and

\int f d

(volume)

= \int^{*} f

.
b. Dual of charge-current 1 -form

J

is charge-current 3 -form

^{*} J

. The total charge

Q

in a 3 -dimensional hypersurface region

S

Q (S) = \int_{S} * J

Box 4.1 (continued)

Conservation of charge is stated locally by

d^{*} J = 0

. Stokes' Theorem goes from this differential conservation law to the integral conservation law,

0 = \int_{q} d^{*} J = \int_{\partial T}^{*} J .

This law is of most interest when

\partial V = S_{2} - S_{1}

consists of the future

S_{2}

and past

S_{1}

boundaries of a spacetime region, in which case it states

Q (S_{2}) =

Q (S_{1})

; see exercise 5.2.
c. Dual of electromagnetic field tensor

F =

Faraday is

* F = M a x w e l l

. From the

d^{*} F = 4 π^{*} J

Maxwell equation, find

4 π Q = 4 π \int_{S}^{*} J = \int_{S} d^{*} F = \int_{\partial S}^{*} F

.
3. Simple forms revisited.
a. The dual of a simple form is simple.
b. Egg crate of

^{*} σ

is perpendicular to egg crate of

σ = α \land β \land \dots \land μ

in this sense:
(1) pick any vector

V

lying in intersection of surfaces of

σ

(⟨ α, V ⟩ = ⟨ β, v ⟩ = \dots = ⟨ μ, V ⟩ = 0)

(2) pick any vector

W

lying in intersection of surfaces of

^{*} σ

;
(3) then

V

and

W

are necessarily perpendicular:

v \cdot w = 0

Example:

σ = 3 d t

is a simple 1 -form in spacetime.
a.

^{*} σ = - 3 d x \land d y \land d z

is a simple 3-form.
b. General vector in surfaces of

σ

V = V^{x} e_{x} + V^{y} e_{y} + V^{z} e_{z} .

c. General vector in intersection of surfaces of

^{*} σ

W = W^{t} e_{t} .

w \cdot V = 0

§4.2. ELECTROMAGNETIC 2-FORM AND LORENTZ FORCE

The electromagnetic field tensor, Faraday

= F

, is an antisymmetric second-rank tensor (i.e., 2 -form). Instead of expanding it in terms of the tensor products of basis 1 -forms,

F = F_{α β} d x^{α} \otimes d x^{β},

the exterior calculus prefers to expand in terms of antisymmetrized tensor products ("exterior products," exercise 4.1):

\begin{aligned} (4.1) & F & = \frac{1}{2} F_{α β} d x^{α} \land d x^{β} \\ (4.2) & d x^{α} \land d x^{β} & \equiv d x^{α} \otimes d x^{β} - d x^{β} \otimes d x^{α} \end{aligned}

Any 2-form (antisymmetric, second-rank tensor) can be so expanded. The symbol "

\land

" is variously called a "wedge," a "hat," or an "exterior product sign"; and

d x^{α} \land d x^{β}

are the "basis 2-forms" of a given Lorentz frame (see §3.5, exercise 3.12, and Box 4.1).

There is no simpler way to illustrate this 2-form representation of the electromagnetic field than to consider a magnetic field in the

x

-direction:

\begin{aligned} (4.3) & F_{y z} & = - F_{z y} = B_{x}, \\ F & = B_{x} d y \land d z . \end{aligned}

The 1 -form

d y = grad y

is the set of surfaces (actually hypersurfaces)

y = 18

(all

t, x, z), y = 19

(all

t, x, z

) ,

y = 20

(all

t, x, z

), etc.; and surfaces uniformly interpolated between them. Similarly for the 1 -form

d z

. The intersection between these two sets of surfaces produces a honeycomb-like structure. That structure becomes a " 2 -form" when it is supplemented by instructions (see arrows in Figure 4.1) that give a "sense of circulation" to each tube of the honeycomb (order of factors in the "wedge product" of equation 4.2;

d y \land d z = - d z \land d y

). The 2-form

F

in the example differs from this "basis 2-form"

d y \land d z

only in this respect, that where

d y \land d z

had one tube, the field 2 -form has

B_{x}

tubes.

When one considers a tubular structure that twists and turns on its way through spacetime, one must have more components to describe it. The 2 -form for the general electromagnetic field can be written as

\begin{aligned} F = & E_{x} d x \land d t + E_{y} d y \land d t + E_{z} d z \land d t + B_{x} d y \land d z \\ (4.4) & + B_{y} d z \land d x + B_{z} d x \land d y \end{aligned}

( 6 components, 6 basis 2 -forms).
A 1 -form is a machine to produce a number out of a vector (bongs of a bell as the vector pierces successive surfaces). A 2 -form is a machine to produce a number out of an oriented surface (surface with a sense of circulation indicated on it: Figure 4.1, lower right). The meaning is as clear here as it is in elementary magnetism:

Electromagnetic 2-form expressed in terms of exterior products

A 2-form as a honeycomb of tubes with a sense of circulation

A 2-form as a machine to produce a number out of an oriented surface

Figure 4.1.
Construction of the 2-form for the electromagnetic field

F = B_{x} d y \land d z

out of the 1-forms

d y

and

d z

by "wedge multiplication" (formation of honeycomb-like structure with sense of circulation indicated by arrows). A 2 -form is a "machine to construct a number out of an oriented surface" (illustrated by sample surface enclosed by arrows at lower right; number of tubes intersected by this surface is

\int_{(this surface)} F = 18;

Faraday's concept of "magnetic flux"). This idea of 2-form machinery can be connected to the "tensor-as-machine" idea of Chapter 3 as follows. The shape of the oriented surface over which one integrates

F

does not matter, for small surfaces. All that affects

\int F

is the area of the surface, and its orientation. Choose two vectors,

u

and

v

, that lie in the surface. They form two legs of a parallelogram, whose orientation (

u

followed by

v

) and area are embodied in the exterior product

u \land v

. Adjust the lengths of

u

and

v

so their parallelogram,

u \land v

, has the same area as the surface of integration. Then

Exercise: derive this result, for an infinitesimal surface

u \land v

and for general

F

, using the formalism of Box 4.1.
the number of Faraday tubes cut by that surface. The electromagnetic 2-form

F

or Faraday described by such a "tubular structure" (suitably abstracted; Box 4.2) has a reality and a location in space that is independent of all coordinate systems and all artificial distinctions between "electric" and "magnetic" fields. Moreover, those tubes provide the most direct geometric representation that anyone has ever been able to give for the machinery by which the electromagnetic field acts on a charged particle. Take a particle of charge

e

and 4 -velocity

\begin{matrix} (4.5) & u = \frac{d x^{α}}{d τ} e_{α} \end{matrix}

Let this particle go through a region where the electromagnetic field is described by the 2 -form

\begin{matrix} (4.6) & F = B_{x} d y \land d z \end{matrix}

of Figure 4.1. Then the force exerted on the particle (regarded as a 1 -form) is the contraction of this 2 -form with the 4 -velocity (and the charge);

\begin{matrix} (4.7) & \dot{ρ} = d p / d τ = e F (u) \equiv e ⟨ F, u ⟩, \end{matrix}

as one sees by direct evaluation, letting the two factors in the 2 -form act in turn on the tangent vector

u

\begin{aligned} \dot{ρ} & = e B_{x} ⟨ d y \land d z, u ⟩ \\ = e B_{x} {d y ⟨ d z, u ⟩ - d z ⟨ d y, u ⟩} \\ = e B_{x} {d y ⟨ d z, u^{z} e_{z} ⟩ - d z ⟨ d y, u^{y} e_{y} ⟩} \end{aligned}

\begin{matrix} (4.8) & {\dot{p}}_{α} d x^{α} = e B_{x} u^{z} d y - e B_{x} u^{y} d z \end{matrix}

Comparing coefficients of the separate basis 1 -forms on the two sides of this equation, one sees reproduced all the detail of the Lorentz force exerted by the magnetic field

B_{x}

\begin{aligned} {\dot{p}}_{y} = \frac{d p_{y}}{d τ} = e B_{x} \frac{d z}{d τ}, \\ (4.9) & {\dot{p}}_{z} = \frac{d p_{z}}{d τ} = - e B_{x} \frac{d y}{d τ} . \end{aligned}

By simple extension of this line of reasoning to the general electromagnetic field, one concludes that the time-rate of change of momentum ( 1 -form) is equal to the charge multiplied by the contraction of the Faraday with the 4-velocity. Figure 4.2 illustrates pictorially how the 2 -form,

F

, serves as a machine to produce the 1 -form,

\dot{p}

, out of the tangent vector, eu.

Box 4.2 ABSTRACTING A 2-FORM FROM THE CONCEPT OF "HONEYCOMBLIKE STRUCTURE," IN 3-SPACE AND IN SPACETIME

Open up a cardboard carton containing a dozen bottles, and observe the honeycomb structure of intersecting north-south and east-west cardboard separators between the bottles. That honeycomb structure of "tubes" ("channels for bottles") is a fairly apt illustration of a 2 -form in the context of everyday 3 -space. It yields a number (number of tubes cut) for each choice of smooth element of 2 -surface slicing through the three-dimensional structure. However, the intersecting cardboard separators are rather too specific. All that a true 2 -form can ever give is the number of tubes sliced through, not the "shape" of the tubes. Slew the carton around on the floor by

45^{\circ}

. Then half the separators run NW-SE and the other half run NE-SW, but through a given bit of 2 -surface fixed in 3 -space the count of tubes is unchanged. Therefore, one should be careful to make the concept of tubes in the mind's eye abstract enough that one envisages direction of tubes (vertical in the example) and density of tubes, but not any specific location or orientation for the tube walls. Thus all the following representations give one and the same 2-form,

σ

\begin{aligned} σ = B d x \land d y \\ σ = B (2 d x) \land (\frac{1}{2} d y) \end{aligned}

(NS cardboards spaced twice as close as before; EW cardboards spaced twice as wide as before);

\begin{matrix} σ = B d (\frac{x - y}{\sqrt{2}}) \land d (\frac{x + y}{\sqrt{2}}) \\ (cardboards rotated through 45) \\ σ = B \frac{α d x + β d y}{(α δ - β γ)^{1 / 2}} \land \frac{γ d x + δ d y}{(α δ - β γ)^{1 / 2}} \end{matrix}

(both orientation and spacing of "cardboards" changing from point to point, with all four
functions,

α, β, γ

, and

δ

, depending on position).

What has physical reality, and constitutes the real geometric object, is not any one of the 1 -forms just encountered individually, but only the 2 -form

σ

itself. This circumstance helps to explain why in the physical literature one sometimes refers to "tubes of force" and sometimes to "lines of force." The two terms for the same structure have this in common, that each yields a number when sliced by a bit of surface. The line-of-force picture has the advantage of not imposing on the mind any specific structure of "sheets of cardboard"; that is, any specific decomposition of the 2 -form into the product of 1 -forms. However, that very feature is also a disadvantage, for in a calculation one often finds it useful to have a well-defined representation of the 2 -form as the wedge product of 1 -forms. Moreover, the tube picture, abstract though it must be if it is to be truthful, also has this advantage, that the orientation of the elementary tubes (sense of circulation as indicated by arrows in Figures 4.1 and 4.5 , for example) lends itself to ready visualization. Let the "walls" of the tubes therefore remain in all pictures drawn in this book as a reminder that 2 -forms can be built out of 1 -forms; but let it be understood here and hereafter how manyfold are the options for the individual 1-forms!
Turn now from three dimensions to four, and find that the concept of "honeycomb-like structure" must be made still more abstract. In three dimensions the arbitrariness of the decomposition of the 2 -form into 1 -forms showed in the slant and packing of the "cardboards," but had no effect on the verticality of the "channels for the bottles" ("direction of Faraday lines of force or tubes of
force"); not so in four dimensions, or at least not in the generic case in four dimensions.

In special cases, the story is almost as simple in four dimensions as in three. An example of a special case is once again the 2 -form

σ = B d x

\land d y

, with all the options for decomposition into 1 -forms that have already been mentioned, but with every option giving the same "direction" for the tubes. If the word "direction" now rises in status from "tube walls unpierced by motion in the direction of increasing

z

" to "tube walls unpierced either by motion in the direction of increasing

z

, or by motion in the direction of increasing

t

, or by any linear combination of such motions," that is a natural enough consequence of adding the new dimension. Moreover, the same simplicity prevails for an electromagnetic plane wave. For example, let the wave be advancing in the

z

-direction, and let the electric polarization point in the

x

-direction; then for a monochromatic wave, one has

E_{x} = B_{y} = E_{0} \cos ω (z - t) = - F_{01} = F_{31},

and all components distinct from these equal zero.
Faraday is

\begin{aligned} F & = F_{01} d t \land d x + F_{31} d z \land d x \\ = E_{0} \cos ω (z - t) d (z - t) \land d x \end{aligned}

which is again representable as a single wedge product of two 1 -forms.

Not so in general! The general 2 -form in four dimensions consists of six distinct wedge products,

\begin{aligned} F = F_{01} d t \land d x + F_{02} d t \land d y & + \dots \\ + F_{23} d y \land d z \end{aligned}

It is too much to hope that this expression will reduce in the generic case to a single wedge product of two 1 -forms ("simple" 2 -form). It is not even
true that it will. It is only remarkable that it can be reduced from six exterior products to two (details in exercise 4.1); thus,

F = π^{1} \land ξ^{1} + π^{2} \land ξ^{2}

Each product

π^{i} \land ξ^{i}

individually can be visualized as a honeycomb-like structure like those depicted in Figures 4.1, 4.2, 4.4, and 4.5. Each such structure individually can be pictured as built out of intersecting sheets (1-forms), but with such details as the tilt and packing of these 1 -forms abstracted away. Each such structure individually gives a number when sliced by an element of surface. What counts for the 2 -form

F

, however, is neither the number of tubes of

π^{1} \land ξ^{1}

cut by the surface, nor the number of tubes of

n^{2} \land ξ^{2}

cut by the surface, but only the sum of the two. This sum is what is referred to in the text as the "number of tubes of

F

" cut by the surface. The contribution of either wedge product individually is not well-defined, for a simple reason: the decomposition of a six-wedge-product object into two wedge products, miraculous though it seems, is actually far from unique (details in exercise 4.2).

In keeping with the need to have two products of 1 -forms to represent the general 2 -form note that the vanishing of

d F

("no magnetic charges") does not automatically imply that

d (n^{1} \land ξ^{1})

d (n^{2} \land ξ^{2})

separately vanish. Note also that any spacelike slice through the general 2 -form

F

(reduction from four dimensions to three) can always be represented in terms of a honeycomb-like structure ("simple" 2 -form in three dimensions; Faraday's picture of magnetic tubes of force).

Despite the abstraction that has gone on in seeing in all generality what a 2 -form is, there is no bar to continuing to use the term "honeycomb-like structure" in a broadened sense to describe this object; and that is the practice here and hereafter.

Figure 4.2.
The Faraday or 2 -form

F

of the electromagnetic field is a machine to produce a 1 -form (the time-rate of change of momentum

\dot{p}

of a charged particle) out of a tangent vector (product of charge

e

of the particle and its 4 -velocity

u

). In spacetime the general 2 -form is the "superposition" (see Box 4.2 ) of two structures like that illustrated at the top of this diagram, the tubes of the first being tilted and packed as indicated, the tubes of the second being tilted in another direction and having a different packing density.

§4.3. FORMS ILLUMINATE ELECTROMAGNETISM, AND ELECTROMAGNETISM ILLUMINATES FORMS

All electromagnetism allows itself to be summarized in the language of 2-forms, honeycomb-like "structures" (again in the abstract sense of "structure" of Box 4.2) of tubes filling all spacetime, as well when spacetime is curved as when it is flat. In brief, there are two such structures, one Faraday

= F

, the other Maxwell

=^{* F}

, each dual ("perpendicular," the only place where metric need enter the discussion) to the other, each satisfying an elementary equation:

\begin{matrix} (4.10) & d F = 0 \end{matrix}

("no tubes of Faraday ever end") and

\begin{matrix} (4.11) & d^{*} F = 4 π^{*} J \end{matrix}

("the number of tubes of Maxwell that end in an elementary volume is equal to the amount of electric charge in that volume"). To see in more detail how this machinery shows up in action, look in turn at: (1) the definition of a 2 -form; (2) the appearance of a given electromagnetic field as Faraday and as Maxwell; (3) the Maxwell structure for a point-charge at rest; (4) the same for a point-charge in motion; (5) the nature of the field of a charge that moves uniformly except during a brief instant of acceleration; (6) the Faraday structure for the field of an oscillating dipole; (7) the concept of exterior derivative; (8) Maxwell's equations in the language of forms; and (9) the solution of Maxwell's equations in flat spacetime, using a 1 -form

A

from which the Liénard-Wiechert 2 -form

F

can be calculated via

F = d A

A 2 -form, as illustrated in Figure 4.1, is a machine to construct a number ("net number of tubes cut") out of any "oriented 2 -surface" (2-surface with "sense of circulation" marked on it):

\begin{matrix} (4.12) & (\begin{array}{l} number \\ of tubes \\ cut \end{array}) = \int_{surface} F . \end{matrix}

For example, let the 2-form be the one illustrated in Figure 4.1

F = B_{x} d y \land d z

and let the surface of integration be the portion of the surface of the 2 -sphere

x^{2} + y^{2} + z^{2} = a^{2}, t =

constant, bounded between

θ = 70^{\circ}

and

θ = 110^{\circ}

and between

φ = 0^{\circ}

and

φ = 90^{\circ}

("Atlantic region of the tropics"). Write

\begin{aligned} y & = a \sin θ \sin φ, \\ z & = a \cos θ, \\ d y & = a (\cos θ \sin φ d θ + \sin θ \cos φ d φ), \\ d z & = - a \sin θ d θ, \\ (4.13) & d y \land d z & = a^{2} \sin^{2} θ \cos φ d θ \land d φ . \end{aligned}

Preview of key points in electromagnetism

A 2-form as machine for number of tubes cut

Number of tubes cut calculated in one example

Figure 4.3.
Spacelike slices through Faraday, the electromagnetic 2-form, a geometric object, a honeycomb of tubes that pervades all spacetime ("honeycomb" in the abstract sense spelled out more precisely in Box 4.2). The surfaces in the drawing do not look like a 2 -form (honeycomb), because the second family of surfaces making up the honeycomb extends in the spatial direction that is suppressed from the drawing. Diagram A shows one spacelike slice through the 2 -form (time increases upwards in the diagram). In diagram B, a projection of the 2 -form on this spacelike hypersurface gives the Faraday tubes of magnetic force in this three-dimensional geometry (if the suppressed dimension were restored, the tubes would be tubes, not channels between lines). Diagram C shows another spacelike slice (hypersurface of simultaneity for an observer in a different Lorentz frame). Diagram D shows the very different pattern of magnetic tubes in this reference system. The demand that magnetic tubes of force shall not end

(\nabla \cdot B = 0)

, repeated over and over for every spacelike slice through Faraday, gives everywhere the result

\partial B / \partial t = - \nabla \times E

. Thus (magnetostatics) + (covariance)

⟶

(magnetodynamics). Similarly-see Chapters 17 and 21(geometrostatics) + (covariance)

⟶

(geometrodynamics).

The structure

d θ \land d θ

looks like a "collapsed egg-crate" (Figure 1.4, upper right) and has zero content, a fact formally evident from the vanishing of

α \land β =

- β \land α

when

α

and

β

are identical. The result of the integration, assuming constant

B_{x}

, is

\begin{matrix} (4.14) & \int_{surface} F = a^{2} B_{x} \int_{70^{\circ}}^{110^{\circ}} \sin^{2} θ d θ \int_{0^{\circ}}^{90^{\circ}} \cos φ d φ \end{matrix}

It is not so easy to visualize a pure electric field by means of its 2 -form

F

(Figure 4.4 , left) as it is to visualize a pure magnetic field by means of its 2 -form

F

(Figures 4.1, 4.2, 4.3). Is there not some way to treat the two fields on more nearly the same footing? Yes, construct the 2 -form *

F

(Figure 4.4, right) that is dual ("perpendicular"; Box 4.3; exercise 3.14) to

F

Figure 4.4.
The Faraday structure

F = \frac{1}{2} F_{μ ν} d x^{μ} \land d x^{y} = \frac{1}{2} F_{01} d t \land d x + \frac{1}{2} F_{10} d x \land d t = E_{x} d x \land d t

associated with an electric field in the

x

-direction, and the dual ("perpendicular") Maxwell honeycomblike 2 -form

^{*} F = \frac{1}{2}^{*} F_{μ ν} d x^{μ} \land d x^{y} =^{*} F_{23} d x^{2} \land d x^{3} = F^{01} d x^{2} \land d x^{3} = F_{10} d x^{2} \land d x^{3} = E_{x} d y \land d z

Represent in geometric form the field of a point-charge of strength

e

at rest at the origin. Operate in flat space with spherical polar coordinates:

\begin{aligned} d s^{2} & = - d τ^{2} = g_{μ ν} d x^{μ} d x^{ν} \\ (4.15) & = - d t^{2} + d r^{2} + r^{2} d θ^{2} + r^{2} \sin^{2} θ d φ^{2} \end{aligned}

The electric field in the

r

-direction being

E_{r} = e / r^{2}

, it follows that the 2 -form

F

or Faraday is

\begin{matrix} (4.16) & F = \frac{1}{2} F_{μ ν} d x^{μ} \land d x^{ν} = - E_{r} d t \land d r = - \frac{e}{r^{2}} d t \land d r \end{matrix}

Its dual, according to the prescription in exercise 3.14, is Maxwell:

\begin{matrix} (4.17) & Maxwell =^{*} F = e \sin θ d θ \land d φ \end{matrix}

Pattern of tubes in dual structure Maxwell for point-charge at rest
as illustrated in Figure 4.5.
Take a tour in the positive sense around a region of the surface of the sphere illustrated in Figure 4.5. The number of tubes of

* F

encompassed in the route will be precisely

(\binom{number}{of tubes}) = e (\binom{solid}{angle})

The whole number of tubes of

* F

emergent over the entire sphere will be

4 π e

, in conformity with Faraday's picture of tubes of force.

Box 4.3 DUALITY OF 2-FORMS IN SPACETIME

Given a general 2-form (containing six exterior or wedge products)

F = E_{x} d x \land d t + E_{y} d y \land d t + \dots + B_{z} d x \land d y

one gets to its dual ("perpendicular") by the prescription

* F = - B_{x} d x \land d t - \dots + E_{y} d z \land d x + E_{z} d x \land d y

Duality Rotations

Note that the dual of the dual is the negative of the original 2 -form; thus

^{* *} F = - E_{x} d x \land d t - \dots - B_{z} d x \land d y = - F

In this sense

^{*}

has the same property as the imaginary number

i :^{* *} = i i = - 1

. Thus one can write

e^{* α} = \cos α +^{*} \sin α

This operation, applied to

F

, carries attention from the generic 2 -form in its simplest representation (see exercise 4.1)

F = E_{x} d x \land d t + B_{x} d y \land d z

to another "duality rotated electromagnetic field"

e^{* α} F = (E_{x} \cos α - B_{x} \sin α) d x \land d t + (B_{x} \cos α + E_{x} \sin α) d y \land d z

If the original field satisfied Maxwell's empty-space field equations, so does the new field. With suitable choice of the "complexion"

α

, one can annul one of the two wedge products at any chosen point in spacetime and have for the other

{(B_{x}^{2} + E_{x}^{2})}^{1 / 2} d y \land d z

Field of a point-charge in motion

How can one determine the structure of tubes associated with a charged particle moving at a uniform velocity? First express *

F

in rectangular coordinates moving with the particle (barred coordinates in this comoving "rocket" frame of reference; unbarred coordinates will be used later for a laboratory frame of reference). The relevant steps can be listed:
(a)

^{*} F = e \sin \bar{θ} d \bar{θ} \land d \bar{φ} = - e (d \cos \bar{θ}) \land d \bar{φ}

Figure 4.5.
The field of 2-forms Maxwell

=^{*} F = e \sin θ d θ \land d ϕ

that describes the electromagnetic field of a charge

e

at rest at the origin. This picture is actually the intersection of

*

with a 3 -surface of constant time

t

; i.e., the time direction is suppressed from the picture.
(b)

\bar{φ} = \arctan \frac{\bar{y}}{\bar{x}}; d \bar{φ} = \frac{\bar{x} d \bar{y} - \bar{y} d \bar{x}}{{\bar{x}}^{2} + {\bar{y}}^{2}};

(c)

\cos \bar{θ} = \frac{\bar{z}}{\bar{r}}; - d (\cos \bar{θ}) = \frac{- d \bar{z}}{\bar{r}} + \frac{\bar{z}}{{\bar{r}}^{3}} (\bar{x} d \bar{x} + \bar{y} d \bar{y} + \bar{z} d \bar{z})

(d) combine to find

\begin{matrix} (4.18) & ^{*} F = (e / {\bar{r}}^{3}) (\bar{x} d \bar{y} \land d \bar{z} + \bar{y} d \bar{z} \land d \bar{x} + \bar{z} d \bar{x} \land d \bar{y}) \end{matrix}

(electromagnetic field of point charge in a comoving Cartesian system; spherically symmetric). Now transform to laboratory coordinates:
velocity parameter

α

velocity

β = \tanh α

\frac{1}{\sqrt{1 - β^{2}}} = \cosh α, \frac{β}{\sqrt{1 - β^{2}}} = \sinh α

(a)

{\begin{cases} \bar{t} = t \cosh α - x \sinh α, \\ \bar{x} = - t \sinh α + x \cosh α, \\ \bar{y} = y \bar{z} = z \end{cases}

(b)

\bar{r} = {[(x \cosh α - t \sinh α)^{2} + y^{2} + z^{2}]}^{1 / 2}

(c)

^{*} F = (e / {\bar{r}}^{3}) [(x \cosh α - t \sinh α) d y \land d z + y d z \land

\begin{matrix} (4.19) & (\cosh α d x - \sinh α d t) + z (\cosh α d x - \sinh α d t) \land d y] \end{matrix}

(d) compare with the general dual 2-form,

\begin{aligned} * F = & E_{x} d y \land d z + E_{y} d z \land d x + E_{z} d x \land d y \\ + B_{x} d t \land d x + B_{y} d t \land d y + B_{z} d t \land d z \end{aligned}

and get the desired individual field components
(e)

{\begin{cases} E_{x} = (e / {\bar{r}}^{3}) (x \cosh α - t \sinh α), & B_{x} = 0, \\ E_{y} = (e / /^{3}) y \cosh α, & B_{y} = - (e / {\bar{r}}^{3}) z \sinh α, \\ E_{z} = (e / r^{3}) z \cosh α, & B_{z} = (e / r^{3}) y \sinh α . \end{cases}

One can verify that the invariants

\begin{aligned} (4.21) & B^{2} - E^{2} = \frac{1}{2} F_{α β} F^{α β}, \\ (4.22) & E \cdot B = \frac{1}{4} F_{α β}^{*} F^{α β} \end{aligned}

have the same value in the laboratory frame as in the rocket frame, as required. Note that the honeycomb structure of the differential form is not changed when one goes from the rocket frame to the laboratory frame. What changes is only the mathematical formula that describes it.

§4.4. RADIATION FIELDS

The Maxwell structure of tubes associated with a charge in uniform motion is more remarkable than it may seem at first sight, and not only because of the Lorentz contraction of the tubes in the direction of motion. The tubes arbitrarily far away move on in military step with the charge on which they center, despite the fact that there is no time for information "emitted" from the charge "right now" to get to the faraway tube "right now." The structure of the faraway tubes "right now" must therefore derive from the charge at an earlier moment on its uniform-motion, straight-line trajectory. This circumstance shows up nowhere more clearly than in what happens to the field in consequence of a sudden change, in a short time

Δ τ

, from one uniform velocity to another uniform velocity (Figure 4.6). The tubes have the standard patterns for the two states of motion, one pattern within a sphere of radius

r

, the other outside that sphere, where

r

is equal to the lapse of time (" cm of light-travel time") since the acceleration took place. The necessity for the two patterns to fit together in the intervening zone, of thickness

Δ r = Δ τ

, forces the field there to be multiplied up by a "stretching factor," proportional to

r

. This factor is responsible for the well-known fact that radiative forces fall off inversely only as the first power of the distance (Figure 4.6).

When the charge continuously changes its state of motion, the structure of the electromagnetic field, though based on the same simple principles as those illustrated in Figure 4.6, nevertheless looks more complex. The following is the Faraday 2-form

Figure 4.6.
Mechanism of radiation. J. J. Thomson's way to understand why the strength of an electromagnetic wave falls only as the inverse first power of distance

r

and why the amplitude of the wave varies (for low velocities) as

\sin θ

(maximum in the plane perpendicular to the line of acceleration). The charge was moving to the left at uniform velocity. Far away from it, the lines of force continue to move as if this uniform velocity were going to continue forever (Coulomb field of point-charge in slow motion). However, closer up the field is that of a point-change moving to the right with uniform velocity (

1 / r^{2}

dependence of strength upon distance). The change from the one field pattern to another is confined to a shell of thickness

Δ τ

located at a distance

r

from the point of acceleration (amplification of field by "stretching factor"

r \sin θ Δ β / Δ τ

; see text). We thank

C

. Teitelboim for the construction of this diagram.
for the field of an electric dipole of magnitude

p_{1}

oscillating up and down parallel
Field of an oscillating dipole to the

z

-axis:

\begin{aligned} F = & E_{x} d x \land d t + \dots + B_{x} d y \land d z + \dots = real part of {p_{1} e^{i ω r - i ω t} \\ \underset{gives E_{r}}{\underset{⏟}{[2 \cos θ (\frac{1}{r^{3}} - \frac{i ω}{r^{2}})}} d r \land d t + \underset{gives E_{θ}}{\underset{⏟}{\sin θ (\frac{1}{r^{3}} - \frac{i ω}{r^{2}} - \frac{ω^{2}}{r})}} r d θ \land d t \\ \underset{gives B_{ϕ}}{\underset{⏟}{\sin θ (\frac{- i ω}{r^{2}} - \frac{ω^{2}}{r})}} d r \land r d θ]} \end{aligned}

and the dual 2 -form Maxwell

=^{*} F

\begin{aligned} * F = & - \underset{gives B_{ϕ}}{\underset{⏟}{- B_{x} d x \land d t - \dots + E_{x} d y \land d z + \dots = real part of {p_{1} e^{i ω r - i ω t}}} \\ \underset{gives E_{r}}{\underset{⏟}{\sin θ (\frac{- i ω}{r^{2}} - \frac{ω^{2}}{r})}} d t \land r \sin θ d ϕ \\ \underset{gives E_{θ}}{\underset{⏟}{2 \cos θ (\frac{1}{r^{3}} - \frac{i ω}{r^{2}})}} r d θ \land r \sin θ d ϕ \\ \underset{r^{3} θ (\frac{1}{r^{2}} - \frac{i ω}{r})}{\underset{⏟}{\sin \sin θ d ϕ \land d r]} .}} \end{aligned}

§4.5. MAXWELL'S EQUATIONS

The general 2 -form

F

is written as a superposition of wedge products with a factor

\frac{1}{2}

\begin{matrix} (4.25) & F = \frac{1}{2} F_{μ ν} d x^{μ} \land d x^{ν} \end{matrix}

because the typical term appears twice, once as

F_{x y} d x \land d y

and the second time as

F_{y x} d y \land d x

, with

F_{y x} = - F_{x y}

and

d y \land d x = - d x \land d y

If differentiation ("taking the gradient"; the operator

d

) produced out of a scalar a 1-form, it is also true that differentiation (again the operator

d

, but now generally known under Cartan's name of "exterior differentiation") produces a 2 -form out of the general 1-form; and applied to a 2 -form produces a 3 -form; and applied to a 3 -form produces a 4 -form, the form of the highest order that spacetime will accommodate. Write the general

f

-form as

\begin{matrix} (4.26) & ϕ = \frac{1}{f!} ϕ_{α_{1} α_{2} \dots α_{f}} d x^{α_{1}} \land d x^{α_{2}} \land \dots \land d x^{α_{f}} \end{matrix}

where the coefficient

ϕ_{α_{1} α_{2} \dots α_{f}}

, like the wedge product that follows it, is antisym-

Taking exterior derivative metric under interchange of any two indices. Then the exterior derivative of

ϕ

\begin{matrix} (4.27) & d ϕ \equiv \frac{1}{f!} \frac{\partial ϕ_{α_{1} α_{2} \dots α_{f}}}{\partial x^{α_{0}}} d x^{α_{0}} \land d x^{α_{1}} \land d x^{α_{2}} \land \dots \land d x^{α_{ϝ}} \end{matrix}

Take the exterior derivative of Faraday according to this rule and find that it vanishes, not only for the special case of the dipole oscillator, but also for a general electromagnetic field. Thus, in the coordinates appropriate for a local Lorentz frame, one has

\begin{aligned} d F = & d (E_{x} d x \land d t + \dots + B_{x} d y \land d z + \dots) \\ = & (\frac{\partial E_{x}}{\partial t} d t + \frac{\partial E_{x}}{\partial x} d x + \frac{\partial E_{x}}{\partial y} d y + \frac{\partial E_{x}}{\partial z} d z) \land d x \land d t \\ (4.28) & + \dots (5 more such sets of 4 terms each) \dots \end{aligned}

Note that such a term as

d y \land d y \land d z

is automatically zero ("collapse of egg-crate cell when stamped on"). Collect the terms that do not vanish and find

\begin{aligned} d F = & (\frac{\partial B_{x}}{\partial x} + \frac{\partial B_{y}}{\partial y} + \frac{\partial B_{z}}{\partial z}) d x \land d y \land d z \\ + (\frac{\partial B_{x}}{\partial t} + \frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z}) d t \land d y \land d z \\ + (\frac{\partial B_{y}}{\partial t} + \frac{\partial E_{x}}{\partial z} - \frac{\partial E_{z}}{\partial x}) d t \land d z \land d x \\ (4.29) & + (\frac{\partial B_{z}}{\partial t} + \frac{\partial E_{y}}{\partial x} - \frac{\partial E_{x}}{\partial y}) d t \land d x \land d y . \end{aligned}

Each term in this expression is familiar from Maxwell's equations

div B = \nabla \cdot B = 0

and

curl E = \nabla \times E = - \dot{B}

Each vanishes, and with their vanishing Faraday itself is seen to have zero exterior derivative:

\begin{matrix} (4.30) & d F = 0 . \end{matrix}

In other words, "Faraday is a closed 2-form"; "the tubes of

F

nowhere come to an end."
A similar calculation gives for the exterior derivative of the dual 2-form Maxwell the result

\begin{aligned} d^{*} F = & d (- B_{x} d x \land d t - \dots + E_{x} d y \land d z + \dots) \\ = & (\frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z}) d x \land d y \land d z \\ + (\frac{\partial E_{x}}{\partial t} - \frac{\partial B_{z}}{\partial y} + \frac{\partial B_{y}}{\partial z}) d t \land d y \land d z \\ + \dots \\ = & 4 π (ρ d x \land d y \land d z \\ - J_{x} d t \land d y \land d z \\ - J_{y} d t \land d z \land d x \\ - J_{z} d t \land d x \land d y) = 4 π^{*} J \\ (4.31) & d^{*} F = & 4 π * J . \end{aligned}

Faraday structure: tubes nowhere end

Maxwell structure: density of tube endings given by charge-current 3 -form

In empty space this exterior derivative, too, vanishes; there Maxwell is a closed 2 -form; the tubes of

^{*} F

, like the tubes of

F

, nowhere come to an end.

In a region where charge is present, the situation changes. Tubes of Maxwell take their origin in such a region. The density of endings is described by the 3 -form *J = charge, a "collection of eggcrate cells" collected along bundles of world lines.

The two equations

d F = 0

and

d^{*} F = 4 π^{*} J

summarize the entire content of Maxwell's equations in geometric language. The forms

F =

Faraday, and

^{*} F = M a x w e l l

, can be described in any coordinates one pleases-or in a language (honeycomb and egg-crate structures) free of any reference whatsoever to coordinates. Remarkably, neither equation makes any reference whatsoever to metric. As Hermann Weyl was one of the most emphatic in stressing (see also Chapters 8 and 9), the concepts of form and exterior derivative are metricfree. Metric made an appearance only in one place, in the concept of duality ("perpendicularity") that carried attention from

F

to the dual structure

^{*} F

§4.6. EXTERIOR DERIVATIVE AND CLOSED FORMS

The words "honeycomb" and "egg crate" may have given some feeling for the geometry that goes with electrodynamics. Now to spell out these concepts more clearly and illustrate in geometric terms, with electrodynamics as subject matter, what it means to speak of "exterior differentiation." Marching around a boundary, yes; but how and why and with what consequences? It is helpful to return to functions and 1 -forms, and see them and the 2 -forms Faraday and Maxwell and the 3 -form charge as part of an ordered progression (see Box 4.4). Two-forms are seen in this box to be of two kinds: (1) a special 2 -form, known as a "closed" 2 -form, which has the property that as many tubes enter a closed 2 -surface as emerge from it (exterior derivative of 2-form zero; no 3-form derivable from it other than the trivial zero 3 -form!); and (2) a general 2 -form, which sends across a closed 2 -surface a non-zero net number of tubes, and therefore permits one to define a nontrivial 3-form ("exterior derivative of the 2 -form"), which has precisely as many egg-crate cells in any closed 2 -surface as the net number of tubes of the 2 -form emerging from that same closed 2 -surface (generalization of Faraday's concept of tubes of force to the world of spacetime, curved as well as flat).

Box 4.4 THE PROGRESSION OF FORMS AND EXTERIOR DERIVATIVES

O-Form or Scalar,

f

An example in the context of 3 -space and Newtonian physics is temperature,

T (x, y, z)

, and in the context of spacetime, a scalar potential,

ϕ (t, x, y, z)

From Scalar to 1-Form

Take the gradient or "exterior derivative" of a scalar

f

to obtain a special 1-form,

γ = d f

. Comments: (a) Any additive constant included in

f

is erased in the process of differentiation; the quantity

n

in the diagram at the left is unknown and irrelevant. (b) The 1 -form

γ

is special in the sense that surfaces in one region "mesh" with surfaces in a neighboring region ("closed 1 -form"). (c) Line integral

\int_{d}^{S / j} d f

is independent of path for any class of paths equivalent to one another under continuous deformation. (d) The 1 -form is a machine to produce a number ("bongs of bell" as each successive integral surface is crossed) out of a displacement (approximation to concept of a tangent vector).

General 1-Form $β = β_{α} d x^{α}$

This is a pattern of surfaces, as illustrated in the diagram at the right; i.e., a machine to produce a number ("bongs of bell";

⟨ β, u ⟩

) out of a vector. A 1-form has a reality and position in space independent of all choice of coordinates. Surfaces do not ordinarily mesh. Integral

\int β

around indicated closed loop does not give zero ("more bongs than antibongs").

From 1-Form to 2-Form

ξ = d β = \frac{\partial β_{α}}{\partial x^{μ}} d x^{μ} \land d x^{α}

ξ

is a pattern of honeycomb-like cells, with a direction of circulation marked on each, so stationed

Box 4.4 (continued)
that the number of cells encompassed in the dotted closed path is identical to the net contribution (excess of bongs over antibongs) for the same path in the diagram of

β

above. The "exterior derivative" is defined so this shall be so; the generalized Stokes theorem codifies it. The word "exterior" comes from the fact that the path goes around the periphery of the region under analysis. Thus the 2 -form is a machine to get a number (number of tubes,

⟨ ξ, u \land v ⟩

) out of a bit of surface

(u \land v)

that has a sense of circulation indicated upon it. The 2 -form thus defined is special in this sense: a rubber sheet "supported around its edges" by

the dotted curve or any other closed curve is crossed by the same number of tubes when: (a) it bulges up in the middle; (b) it is pushed down in the middle; (c) it experiences any other continuous deformation. The Faraday or 2 -form

F

of electromagnetism, always expressible as

F = d A

(

A = 4

-potential, a 1 -form), also has always this special property ("conservation of tubes").

O-Form to $1$ -Form to 2 -Form? No!

Go from scalar

f

to 1 -form

γ = d f

. The next step to a 2 -form

α

is vacuous. The net contribution of the line integral

\int y

around the dotted closed path is automatically zero. To reproduce that zero result requires a zero 2 -form. Thus

α = d γ = d d f

has to be the zero 2 -form. This result is a special instance of the general result

d d = 0

General 2-Form

σ = \frac{1}{2} σ_{α β} d x^{α} \land d x^{β}

, with

σ_{α β} = - σ_{β α}

Again, this is a honeycomb-like structure, and again a machine to get a number (number of tubes,

⟨ σ, u \land v ⟩

) out of a surface (

u \land v

) that has a sense of circulation indicated on it. It is general in the sense that the honeycomb structures in one region do not ordinarily mesh with those
in a neighboring region. In consequence, a closed 2 -surface, such as the box-like surface indicated by dotted lines at the right, is ordinarily crossed by a non-zero net number of tubes. The net number of tubes emerging from such a closed surface is, however, exactly zero when the 2 -form is the exterior derivative of a 1 -form.

From 2-Form to 3-Form

μ = d σ = \frac{\partial σ_{| α β |}}{\partial x^{γ}} d x^{γ} \land d x^{α} \land d x^{β}

,
where

d x^{γ} \land d x^{α} \land d x^{β} \equiv 3! d x^{[γ} \otimes d x^{α} \otimes d x^{β]}

This egg-crate type of structure is a machine to get a number (number of cells

⟨ μ, u \land v \land w ⟩

) from a volume (volume

u \land v \land w

within which one counts the cells). A more complete diagram would provide each cell and the volume of integration itself with an indicator of orientation (analogous to the arrow of circulation shown for cells of the 2 -form). The contribution of a given cell to the count of cells is +1 or -1 , according as the orientation indicators have same sense or opposite sense. The number of egg-crate cells of

μ = d σ

in any given volume (such as the volume indicated by the dotted lines) is tailored to give precisely the same number as the net number of tubes of the 2 -form

σ

(diagram above) that emerge

from that volume (generalized Stokes theorem). For electromagnetism, the exterior derivative of Faraday or 2 -form

F

gives a null 3 -form, but the exterior derivative of Maxwell or 2 -form *F gives

4 π

times the 3 -form

* J

of charge:

\begin{aligned} * J = ρ d & x \land d y \land d z - J_{x} d t \land d y \land d z \\ - J_{y} d t \land d z \land d x - J_{z} d t \land d x \land d y \end{aligned}

Box 4.4 (continued)

From 1-Form to 2-Form to 3-Form? No!

Starting with a 1 -form (electromagnetic 4-potential), construct its exterior derivative, the 2 -form

F = d A

(Faraday). The tubes in this honeycomb-like structure never end. So the number of tube endings in any elementary volume, and with it the 3 -form

d F = d d A

, is automatically zero. This is another example of the general result that

d d = 0

From 2-Form to 3-Form to 4-Form? No!

Starting with 2 -form *

F

(Maxwell), construct its exterior derivative, the 3-form

4 π * J

. The cells in this egg-crate type of structure extend in a fourth dimension ("hypertube"). The number of these hypertubes that end in any elementary 4 -volume, and with it the 4 -form

d (4 π^{*} J) = d d^{*} F

is automatically zero, still another example of the general result that

d d = 0

. This result says that

d^{*} J = (\frac{\partial ρ}{\partial t} + \frac{\partial J_{x}}{\partial x} + \frac{\partial J_{y}}{\partial y} + \frac{\partial J_{z}}{\partial z}) d t \land d x \land d y \land d z = 0

("law of conservation of charge"). Note:

d x^{α} \land d x^{β} \land d x^{γ} \land d x^{δ} \equiv 4! d x^{[α} \otimes d x^{β} \otimes d x^{γ} \otimes d x^{δ]}

This implies

d t \land d x \land d y \land d z = ε

From 3-Form to 4-Form

τ = d ν = \frac{\partial ν_{| α β γ |}}{\partial x^{δ}} d x^{δ} \land d x^{α} \land d x^{β} \land d x^{γ}

This four-dimensional "super-egg-crate" type structure is a machine to get a number (number of cells,

⟨ r, n \land u \land v \land w ⟩

) from a 4-volume

n \land u \land v \land w

From 4-Form to 5-Form? No!

Spacetime, being four-dimensional, cannot accommodate five-dimensional egg-crate structures. At least two of the

d x^{μ}

's in

d x^{α} \land d x^{β} \land d x^{γ} \land d x^{δ} \land d x^{ε}

must be the same; so, by antisymmetry of "

\land

," this "basis 5 -form" must vanish.

Results of Exterior Differentiation, Summarized

0-form	$f$
1-form	$d f$	$A$
2-form	$d d f \equiv 0$	$F = d A$	$^{*} F$
3-form		$d F = d d A \equiv 0$	$4 π^{} J = d^{} F$	$ν$
4-form			$d (4 π^{} J) = d d^{} F \equiv 0$	$τ = d ν$	$μ$
5-form?	No!			$d m \equiv 0$	$d μ \equiv 0$

New Forms from Old by Taking Dual (see exercise 3.14)

Dual of scalar

f

is 4-form:

^{*} f = f d x^{0} \land d x^{1} \land d x^{2} \land d x^{3} = f ε

.
Dual of 1-form

J

is 3-form:

^{*} J = J^{0} d x^{1} \land d x^{2} \land d x^{3} - J^{1} d x^{2} \land d x^{3} \land d x^{0}

+ J^{2} d x^{3} \land d x^{0} \land d x^{1} - J^{3} d x^{0} \land d x^{1} \land d x^{2}

.
Dual of 2-form

F

is 2-form:

^{*} F = F^{| α β |} ε_{α β | μ ν |} d x^{μ} \land d x^{p}

, where

F^{α β} = η^{α λ} η^{β δ} F_{λ δ}

.
Dual of 3-form

K

is 1-form:

^{*} K = K^{012} d x^{3} - K^{123} d x^{0} + K^{230} d x^{1} - K^{301} d x^{2}

, where

K^{α β γ} = η^{α μ} η^{β^{ν} η^{γ λ}} K_{μ ν λ}

.
Dual of 4-form

L

is a scalar:

L = L_{0123} d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}

;

^{*} L = L^{0123} = - L_{0123} .

Note 1: This concept of duality between one form and another is to be distinguished from the concept of duality between the vector basis

e_{α}

and the 1-form basis

ω^{α}

of a given frame. The two types of duality have nothing whatsoever to do with each other!

Box 4.4 (continued)

Note 2: In spacetime, the operation of taking the dual, applied twice, leads back to the original form for forms of odd order, and to the negative thereof for forms of even order. In Euclidean 3-space the operation reproduces the original form, regardless of its order.

Duality Plus Exterior Differentiation

Start with scalar

ϕ

. Its gradient

d ϕ

is a 1 -form. Take its dual, to get the 3 -form *

d ϕ

. Take its exterior derivative, to get the 4 -form

d^{*} d ϕ

. Take its dual, to get the scalar

◻ ϕ \equiv -^{*} d * d d

. Verify by index manipulations that

◻

as defined here is the wave operator; i.e., in any Lorentz frame,

◻ ϕ = ϕ_{, α^{, α}} = - (\partial^{2} ϕ / \partial t^{2}) + \nabla^{2} ϕ

Start with 1 -form

A

. Get 2 -form

F = d A

. Take its dual

^{*} F =^{*} d A

, also a 2 -form. Take its exterior derivative, obtaining the 3 -form

d^{*} F

(has value

4 π^{*} J

in electromagnetism). Take its dual, obtaining the 1 -form

^{*} d^{*} F =^{*} d d^{*} d A = 4 π J

("Wave equation for electromagnetic 4-potential"). Reduce in index notation to

F_{μ ν}^{, ν} = A_{ν, μ}^{, ν} - A_{μ, v^{, ν}} = 4 π J_{μ} .

[More in Flanders (1963) or Misner and Wheeler (1957); see also exercise 3.17.]

§4.7. DISTANT ACTION FROM LOCAL LAW

Differential forms are a powerful tool in electromagnetic theory, but full power requires mastery of other tools as well. Action-at-a-distance techniques ("Green's functions," "propagators") are of special importance. Moreover, the passage from Maxwell field equations to electromagnetic action at a distance provides a preview of how Einstein's local equations will reproduce (approximately) Newton's

1 / r^{2}

law.

In flat spacetime and in a Lorentz coordinate system, express the coordinates of particle

A

as a function of its proper time

α

, thus:

\begin{matrix} (4.32) & a^{μ} = a^{μ} (α), \frac{d a^{μ}}{d α} = {\dot{a}}^{μ} (α), \frac{d^{2} a^{μ}}{d α^{2}} = {\ddot{a}}^{μ} (α) . \end{matrix}

Dirac found it helpful to express the distribution of charge and current for a particle of charge

e

following such a motion as a superposition of charges that momentarily
flash into existence and then flash out of existence. Any such flash has a localization in space and time that can be written as the product of four Dirac delta functions [see, for example, Schwartz (1950-1951), Lighthill (1958)]:

\begin{matrix} (4.33) & δ^{4} (x^{μ} - a^{μ}) = δ [x^{0} - a^{0} (α)] δ [x^{1} - a^{1} (α)] δ [x^{2} - a^{2} (α)] δ [x^{3} - a^{3} (α)] . \end{matrix}

Here any single Dirac function

δ (x)

("symbolic function"; "distribution"; "limit of a Gauss error function" as width is made indefinitely narrow and peak indefinitely high, with integrated value always unity) both (1) vanishes for

x \neq 0

, and (2) has the integral

\int_{- \infty}^{+ \infty} δ (x) d x = 1

. Described in these terms, the density-current vector for the particle has the value ("superposition of flashes")

\begin{matrix} (4.34) & J^{μ} = e \int δ^{4} [x^{ν} - a^{ν} (α)] {\dot{a}}^{μ} (α) d α \end{matrix}

The density-current (4.34) drives the electromagnetic field,

F

. Write

F = d A

to satisfy automatically half of Maxwell's equations

(d F = d d A \equiv 0)

\begin{matrix} (4.35) & F_{μ α} = \frac{\partial A_{α}}{\partial x^{μ}} - \frac{\partial A_{μ}}{\partial x^{α}} . \end{matrix}

In flat space, the remainder of Maxwell's equations

(d^{*} F = 4 π^{*} J)

become

\frac{\partial F_{μ}^{v}}{\partial x^{v}} = 4 π J_{μ}

\begin{matrix} (4.36) & \frac{\partial}{\partial x^{μ}} \frac{\partial A^{ν}}{\partial x^{ν}} - η^{ν α} \frac{\partial^{2} A_{μ}}{\partial x^{ν} \partial x^{α}} = 4 π J_{μ} . \end{matrix}

Make use of the freedom that exists in the choice of 4-potentials

A^{ν}

to demand

\begin{matrix} (4.37) & \frac{\partial A^{v}}{\partial x^{y}} = 0 \end{matrix}

(Lorentz gauge condition; see exercise 3.17). Thus get

\begin{matrix} (4.38) & 1 A_{μ} = - 4 π J_{μ} \end{matrix}

The density-current being the superposition of "flashes," the effect

(A)

of this density-current can be expressed as the superposition of the effects

E

of elementary flashes; thus

\begin{matrix} (4.39) & A^{μ} (x) = \int E [x - a (α)] {\dot{a}}^{μ} (α) d α \end{matrix}

where the "elementary effect"

E

("kernel"; "Green's function") satisfies the equation

\begin{matrix} (4.40) & ◻ E (x) = - 4 π δ^{4} (x) . \end{matrix}

One solution is the "half-advanced-plus-half-retarded potential,"

\begin{matrix} (4.41) & E (x) = δ (η_{α β} x^{α} x^{β}) . \end{matrix}

World line of charge regarded as succession of flash-on, flash-off charges

It vanishes everywhere except on the backward and forward light cones, where it has equal strength. Normally more useful is the retarded solution,

\begin{matrix} (4.42) & R (x) = {\begin{cases} 2 E (x) & if x^{0} > 0, \\ 0 & if x^{0} < 0 \end{cases} \end{matrix}

which is obtained by doubling (4.41) in the region of the forward light cone and nullifying it in the region of the backward light cone. All electrodynamics (Coulomb forces, Ampère's law, electromagnetic induction, radiation) follows from the simple expression (4.39) for the vector potential [see, e.g., Wheeler and Feynman (1945) and (1949), also Rohrlich (1965)].

EXERCISES

Exercise 4.1. GENERIC LOCAL ELECTROMAGNETIC FIELD EXPRESSED IN SIMPLEST FORM

In the laboratory Lorentz frame, the electric field is

E

, the magnetic field

B

. Special cases are: (1) pure electric field

(B = 0)

; (2) pure magnetic field (

E = 0

); and (3) "radiation field" or "null field" (

E

and

B

equal in magnitude and perpendicular in direction). All cases other than (1), (2), and (3) are "generic." In the generic case, calculate the Poynting density of flow of energy

E \times B / 4 π

and the density of energy

(E^{2} + B^{2}) / 8 π

. Define the direction of a unit vector

n

and the magnitude of a velocity parameter

α

by the ratio of energy flow to energy density:

n \tanh 2 α = \frac{2 E \times B}{E^{2} + B^{2}} .

View the same electromagnetic field in a rocket frame moving in the direction of

n

with the velocity parameter

α

(not

2 α

; factor 2 comes in because energy flow and energy density are components, not of a vector, but of a tensor). By employing the formulas for a Lorentz transformation (equation 3.23), or otherwise, show that the energy flux vanishes in the rocket frame, with the consequence that

\overset{―}{E}

and

\overset{―}{B}

are parallel. No one can prevent the

\bar{z}

-axis from being put in the direction common to

\overset{―}{E}

and

\overset{―}{B}

. Show that with this choice of direction, Faraday becomes

F = {\bar{E}}_{z} d \bar{z} \land d \bar{t} + {\bar{B}}_{z} d \bar{x} \land d \bar{y}

(only two wedge products needed to represent the generic local field; "canonical representation"; valid in one frame, valid in any frame).

Exercise 4.2. FREEDOM OF CHOICE OF 1-FORMS IN CANONICAL representation of generic local field

Deal with a region so small that the variation of the field from place to place can be neglected. Write Faraday in canonical representation in the form

F = d p_{I} \land d q^{I} + d p_{I I} \land d q^{I I}

where

p_{A} (A = I

I I)

and

q^{A}

are scalar functions of position in spacetime. Define a "canonical transformation" to new scalar functions of position

p_{\bar{A}}

and

q^{\bar{A}}

by way of the "equation of transformation"

p_{A} d q^{A} = d S + p_{\bar{A}} d q^{\bar{A}},

Figure 4.7.
Some simple types of 1 -forms compared and contrasted.
where the "generating function"

S

of the transformation is an arbitrary function of the

q^{A}

and the

q^{\bar{A}}

d S = (\partial S / \partial q^{A}) d q^{A} + (\partial S / \partial q^{\bar{A}}) d q^{\bar{A}}

(a) Derive expressions for the two

p_{A}

's and the two

p_{\bar{A}}

's in terms of

S

by equating coefficients of

d q^{I}, d q^{I I}, d q^{I}, d q^{I I}

individually on the two sides of the equation of transformation.
(b) Use these expressions for the

p_{A}

's and

p_{\bar{A}}

's to show that

F = d p_{A} \land d q^{A}

and

\overset{―}{F} =

d p_{\bar{A}} \land d q^{\bar{A}}

, ostensibly different, are actually expressions for one and the same 2 -form in terms of alternative sets of 1 -forms.

Exercise 4.3. A CLOSED OR CURL-FREE 1-FORM IS A GRADIENT

Given a 1 -form

σ

such that

d σ = 0

, show that

σ

can be expressed in the form

σ = d f

, where

f

is some scalar. The 1 -form

σ

is said to be "curl-free," a narrower category of 1 -form than the "rotation-free" 1-form of the next exercise (expressible as

σ = h d f

), and it in turn is narrower (see Figure 4.7) than the category of " 1 -forms with rotation" (not expressible in the form

σ = h d f

). When the 1 -form

σ

is expressed in terms of basis 1-forms

d x^{α}

, multiplied by corresponding components

σ_{α}

, show that "curl-free" implies

σ_{[α, β]} = 0

Exercise 4.4. CANONICAL EXPRESSION FOR A ROTATION-FREE 1-FORM

In three dimensions a rigid body turning with angular velocity

ω

about the

z

-axis has components of velocity

v_{y} = ω x

, and

v_{x} = - ω y

. The quantity curl

v = \nabla \times v

has

z

-component equal to

2 ω

, and all other components equal to zero. Thus the scalar product of

v

and curl

v

vanishes:

v_{[i, j} v_{k]} = 0

The same concept generalizes to four dimensions,

v_{[α, β} v_{γ]} = 0

and lends itself to expression in coordinate-free language, as the requirement that a certain 3 -form must vanish:

d v \land v = 0

Any 1-form

v

satisfying this condition is said to be "rotation-free." Show that a 1 -form is rotation-free if and only if it can be written in the form

v = h d f

where

h

and

f

are scalar functions of position (the "Frobenius theorem").

Exercise 4.5. FORMS ENDOWED WITH POLAR SINGULARITIES

List the principal results on how such forms are representable, such as

Φ_{1} = \frac{d S}{S} \land ψ_{1} + θ_{1}

and the conditions under which each applies [for the meaning and answer to this exercise, see Lascoux (1968)].

Exercise 4.6. THE FIELD OF THE OSCILLATING DIPOLE

Verify that the expressions given for the electromagnetic field of an oscillating dipole in equations (4.23) and (4.24) satisfy

d F = 0

everywhere and

d^{*} F = 0

everywhere except at the origin.

Exercise 4.7. THE 2-FORM MACHINERY TRANSLATED INTO TENSOR MACHINERY

This exercise is stated at the end of the legend caption of Figure 4.1.

Exercise 4.8. PANCAKING THE COULOMB FIELD

Figure 4.5 shows a spacelike slice,

t =

const, through the Maxwell of a point-charge at rest. By the following pictorial steps, verify that the electric-field lines get compressed into the transverse direction when viewed from a moving Lorentz frame: (1) Draw a picture of an equatorial slice (

θ = π / 2; t, r, ϕ

variable) through Maxwell

=^{*} F

. (2) Draw various spacelike slices, corresponding to constant time in various Lorentz frames, through the resultant geometric structure. (3) Interpret the intersection of Maxwell

= * F

with each Lorentz slice in the manner of Figure 4.3.

Exercise 4.9. COMPUTATION OF SURFACE INTEGRALS

In Box 4.1 the definition

\int α = \int \dots \int ⟨ α, \frac{\partial P}{\partial λ^{1}} \land \dots \land \frac{\partial P}{\partial λ^{p}} ⟩ d λ^{1} \dots d λ^{p}

is given for the integral of a

p

-form

α

over a

p

-surface

P (λ^{1}, \dots, λ^{p})

n

-dimensional space. From this show that the following computational rule (also given in Box 4.1) works: (1) substitute the equation for the surface,

x^{k} = x^{k} (λ^{1}, \dots, λ^{p})

into

α

and collect terms in the form

α = a (λ^{1}, \dots, λ^{p}) d λ^{1} \land \dots \land d λ^{p}

(2) integrate

\int α = \int \dots \int a (λ^{1}, \dots, λ^{p}) d λ^{1} \dots d λ^{p}

using the elementary definition of integration.

Exercise 4.10. WHITAKER'S CALUMOID, OR, THE LIFE OF A LOOP

Take a closed loop, bounding a 2 -dimensional surface

S

. It entraps a certain flux of Faraday

Φ_{F} = \int_{S} F

("magnetic tubes") and a certain flux of Maxwell

Φ_{M} = \int_{S} * F

("electric tubes").
(a) Show that the fluxes

Φ_{F}

and

Φ_{M}

depend only on the choice of loop, and not on the choice of the surface

S

bounded by the loop, if and only if

d F = d^{*} F = 0

(no magnetic charge; no electric charge). Hint: use generalized Stokes theorem, Boxes 4.1 and 4.6.
(b) Move the loop in space and time so that it continues to entrap the same two fluxes. Move it forward a little more here, a little less there, so that it continues to do so. In this way trace out a 2-dimensional surface ("calumoid"; see E. T. Whitaker 1904)

P = P (a, b)

;

x^{μ} = x^{μ} (a, b)

. Show that the elementary bivector in this surface,

Σ = \partial P / \partial a \land \partial P / \partial b

satisfies

⟨ F, Σ ⟩ = 0

and

⟨ * F, Σ ⟩ = 0

.
(c) Show that these differential equations for

x^{μ} (a, b)

can possess a solution, with given initial condition

x^{μ} = x^{μ} (a, 0)

for the initial location of the loop, if

d F = 0

and

d^{*} F = 0

(no magnetic charge, no electric charge).
(d) Consider a static, uniform electric field

F = - E_{x} d t \land d x

. Solve the equations,

⟨ F, Σ ⟩ = 0

and

⟨^{*} F, Σ ⟩ = 0

to find the equation

P (a, b)

for the most general calumoid. [Answer:

y = y (a), z = z (a), x = x (b), t = t (b)

.] Exhibit two special cases: (i) a calumoid that lies entirely in a hypersurface of constant time [loop moves at infinite velocity; analogous to super-light velocity of point of crossing for two blades of a pair of scissors]; (ii) a calumoid whose loop remains forever at rest in the

t, x, y, z

Lorentz frame.

Exercise 4.11. DIFFERENTIAL FORMS AND HAMILTONIAN MECHANICS

Consider a dynamic system endowed with two degrees of freedom. For the definition of this system as a Hamiltonian system (special case: here the Hamiltonian is independent of time), one needs (1) a definition of canonical variables (see Box 4.5) and (2) a knowledge of the Hamiltonian

H

as a function of the coordinates

q^{1}, q^{2}

and the canonically conjugate momenta

p_{1}, p_{2}

. To derive the laws of mechanics, consider the five-dimensional space of

p_{1}, p_{2}, q^{1}, q^{2}

, and

t

, and a curve in this space leading from starting values of the five coordinates (subscript

A

) to final values (subscript

B

), and the value

I = \int_{A}^{B} p_{1} d q^{1} + p_{2} d q^{2} - H (p, q) d t = \int_{A}^{B} ω

of the integral

I

taken along this path. The difference of the integral for two "neighboring" paths enclosing a two-dimensional region

S

, according to the theorem of Stokes (Boxes 4.1 and 4.6), is

δ I = f_{r} ω = \int_{s} d ω

The principle of least action (principle of "extremal history") states that the representative point of the system must travel along a route in the five-dimensional manifold (route with tangent vector

d P / d t)

such that the variation vanishes for this path; i.e.,

d ω (\dots, d P / d t) = 0

(2-form

d ω

with a single vector argument supplied, and other slot left unfilled, gives the 1 -form in 5-space that must vanish). This fixes only the direction of

d P / d t

; its magnitude can be normalized by requiring

⟨ d t, d P / d t ⟩ = 1

.
(a) Evaluate

d ω

from the expression

ω = p_{j} d q^{j} - H d t

.
(b) Set

d P / d t = {\dot{q}}^{j} (\partial P / \partial q^{j}) + {\dot{p}}_{j} (\partial P / \partial p_{j}) + \dot{i} (\partial P / \partial t)

, and expand

d ω (\dots, d P / d t) = 0

in terms of the basis

{d p_{j}, d q^{k}, d t}

Box 4.5 METRIC STRUCTURE AND HAMILTONIAN OR "SYMPLECTIC STRUCTURE" COMPARED AND CONTRASTED

	Metric structure	Symplectic structure
1. Physical application	Geometry of spacetime	Hamiltonian mechanics
2. Canonical structure	$\begin{aligned} (\dots \cdot) = " d s^{2} " = - d t \otimes d t \\ + d x \otimes d x + d y \otimes d y \\ + d z \otimes d z \end{aligned}$	$O = d p_{1} \land d q^{1} + d p_{2} \land d q^{2}$
3. Nature of "metric"	Symmetric	Antisymmetric
4. Name for given coordinate system and any other set of four coordinates in which metric has same form	Lorentz coordinate system	System of "canonically" (or "dynamically") conjugate coordinates
5. Field equation for this metric	$R_{μ ν α β} = 0$ (zero Riemann curvature; flat spacetime)	$\begin{aligned} d O = 0 ("closed 2-form"; \\ condition automatically \\ satisfied by expression above). \end{aligned}$
6. The four-dimensional manifold	Spacetime	Phase space
7. Coordinate-free description of the structure of this manifold	$R i e m a n n = 0$	$d O = 0$
8. Canonical coordinates distinguished from other	Make metric take above form (item 2)	Make metric take above form (item 2)

{q^{j}, p_{k}}

O (\dots, d P / d t) = d H

where

O

is the 2 -form defined in Box 4.5.
(d) Show that the components of

O (\dots, d P / d t) = d H

in the

{q^{j}, p_{k}}

coordinate system are the familiar Hamilton equations. Note that this conclusion depends only on the form assumed for

Θ

, so that one also obtains the standard Hamilton equations in any other phase-space coordinates

{{\bar{q}}^{j}, {\bar{p}}_{k}}

("canonical variables") for which

O = d {\bar{p}}_{1} \land d {\bar{q}}^{1} + d \overset{―}{p_{2}} \land d {\bar{q}}^{2}

Exercise 4.12. SYMMETRY OPERATIONS AS TENSORS

We define the meaning of square and round brackets enclosing a set of indices as follows:

V_{(α_{1} \dots α_{p})} \equiv \frac{1}{p!} Σ V_{α_{α_{1}, \dots} \dots α_{π p}}, V_{[α_{1} \dots α_{p}]} \equiv \frac{1}{p!} Σ (- 1)^{π} V_{α_{π i} \dots α_{π,}}

Box 4.6 BIRTH OF STOKES' THEOREM

Central to the mathematical formulation of electromagnetism are the theorems of Gauss (taken up in Chapter 5) and Stokes. Both today appear together as one unity when expressed in the language of forms. In earlier times the unity was not evident. Everitt (1970) recalls the history of Stokes' theorem: "The Smith's Prize paper set by [G. C.] Stokes [Lucasian Professor of Mathematics] and taken by Maxwell in [February] 1854 ...
5. Given the centre and two points of an ellipse, and the length of the major axis, find its direction by a geometrical construction.
6. Integrate the differential equation

(a^{2} - x^{2}) d y^{2} + 2 x y d y d x + (a^{2} - y^{2}) d x^{2} = 0 .

Has it a singular solution?
7. In a double system of curves of double curvature, a tangent is always drawn at the variable point

P

; shew that, as

P

moves away from an arbitrary fixed point

Q

, it must begin to move along a generating line of an elliptic cone having

Q

for vertex in order that consecutive tangents may ultimately intersect, but that the conditions of the problem may be impossible.
8. If

X, Y, Z

be functions of the rectangular co-ordinates

x, y, z, d S

an element of any limited surface,

l, m, n

the cosines of the inclinations of the normal at

d S

to the axes,

d s

an element of the bounding line, shew that

\begin{matrix} \iint {l (\frac{d Z}{d y} - \frac{d Y}{d z}) + m (\frac{d X}{d z} - \frac{d Z}{d x}) + n (\frac{d Y}{d x} - \frac{d X}{d y})} d S \\ = \int (X \frac{d x}{d s} + Y \frac{d y}{d s} + Z \frac{d z}{d s}) d s \end{matrix}

the differential coefficients of

X, Y, Z

being partial, and the single integral being taken all round the perimeter of the surface
marks the first appearance in print of the formula connecting line and surface integrals now known as Stokes' theorem. This was of great importance to Maxwell's development of electromagnetic theory. The earliest explicit proof of the theorem appears to be that given in a letter from Thomson to Stokes dated July 2, 1850." [Quoted in Campbell and Garnett (1882), pp. 186-187.]

Here the sum is taken over all permutations

π

of the numbers

1, 2, \dots, p

, and

(- 1)^{π}

is +1 or -1 depending on whether the permutation is even or odd. The quantity

V

may have other indices, not shown here, besides the set of

p

indices

α_{1}, α_{2}, \dots, α_{p}

, but only this set of indices is affected by the operations described here. The numbers

π_{1}, π_{2}, \dots, π_{p}

are the numbers

1, 2, \dots, p

rearranged according to the permutation

π

. (Cases

p = 2, 3

were treated in exercise 3.12.) We therefore have machinery to convert any rank-p tensor with components

V_{α_{1} \dots α_{p}}

into a new tensor with components

[A / t (V)]_{μ_{1} \dots μ_{p}} = V_{[μ_{1} \dots μ_{p}]} .

Since this machinery

A / t

is linear, it can be viewed as a tensor which, given suitable arguments

u, v, \dots, w, α, β, \dots, γ

produces a number

u^{μ} v^{ν} \dots w^{λ} α_{[μ} β_{ν} \dots γ_{λ]}

(a) Show that the components of this tensor are

(A I)_{β_{1} \dots β_{p}}^{α_{1} \dots α_{p}} = (p!)^{- 1} δ_{β_{1} \dots β_{p}}^{α_{1} \dots α_{p}}

(Note: indices of

δ

are almost never raised or lowered, so this notation leads to no confusion.)
where

δ_{β_{1} \dots β_{p}}^{α_{1} \dots α_{p}} = {\begin{cases} + 1 if (α_{1}, \dots, α_{p}) is an even permutation of (β_{1}, \dots, β_{p}), \\ - 1 if (α_{1}, \dots, α_{p}) is an odd permutation of (β_{1}, \dots, β_{p}), \\ 0 if (i) any two of the α's are the same, \\ 0 if (ii) any two of the β's are the same, \\ 0 if (iii) the α's and β's are different sets of integers. \end{cases}

Note that the demonstration, and therefore these component values, are correct in any frame.
(b) Show for any "alternating" (i.e., "completely antisymmetric") tensor

A_{α_{1} \dots α_{p}} = A_{[α_{1} \dots α_{p}]}

that

\begin{aligned} \frac{1}{p!} A_{α_{1} \dots α_{p}} & δ_{γ_{1} \dots γ_{p} γ_{p + 1} \dots γ_{p + q}}^{α_{1} \dots α_{p} β_{1} \dots β_{q}} \\ = \sum_{α_{1} < α_{2} < \dots < α_{p}} A_{α_{1} \dots α_{p}} δ_{γ_{1} \dots \dots p \dots γ_{p + q}}^{α_{1} \dots α_{p} β_{1} \dots β_{q}} \\ \equiv A_{| α_{1} \dots α_{p ∣} |} δ_{γ_{1} \dots \dots \dots \dots γ_{p + q}}^{α_{1} \dots α_{p} \dots β_{q}} \end{aligned}

The final line here introduces the convention that a summation over indices enclosed between vertical bars includes only terms with those indices in increasing order. Show, consequently or similarly, that

δ_{γ_{1} \dots \dots \dots \dots γ_{p + q}}^{α_{1} \dots α_{p} β_{1} \dots β_{q}} δ_{| β_{1} \dots β_{q} |}^{μ_{1} \dots μ_{q}} = δ_{γ_{1} \dots \dots \dots \dots γ_{p + q}}^{α_{1} \dots α_{p} μ_{1} \dots μ_{q}} .

(α \land β)_{λ_{1} \dots λ_{p + q}} = δ_{λ_{1} \dots λ_{p} λ_{p + 1} \dots λ_{p + q}}^{μ_{1} \dots μ_{p} ν_{1} \dots ν_{q}} α_{| μ_{1} \dots μ_{p} |} β_{| ν_{1} \dots ν_{q} |};

and similarly

(U \land V)^{λ_{1} \dots λ_{p + q}} = δ_{μ_{1} \dots μ_{p} p_{1} \dots v_{q}}^{λ_{1} \dots λ_{p} λ_{p 1 +} \dots λ_{p + q}} U^{| μ_{1} \dots μ_{p} |} V^{| p_{1} \dots ν_{q} |}

Show that this implies equation (3.45b). Establish the associative law for this product rule by showing that

\begin{aligned} [(α \land β) \land γ]_{σ_{1} \dots σ_{p + q + r}} \end{aligned}

\begin{aligned} = [α \land (β \land γ)]_{σ_{1} \dots σ_{p + α + r}}; \end{aligned}

and show that this reduces to the 3-form version of Equation (3.45c) when

α, β

, and

γ

are all 1 -forms.
(d) Derive the following formula for the components of the exterior product of

p

vectors

\begin{aligned} {(u_{1} \land u_{2} \land \dots \land u_{p})}^{α_{1} \dots α_{p}} & = δ_{μ \dots ν}^{α_{1} \dots α_{p}} {(u_{1})}^{μ} \dots {(u_{p})}^{v} \\ = p! u_{1}^{[α_{1}} u_{2}^{α_{2}} \dots u_{p}^{α_{p}]} \\ = δ_{12 \dots p}^{α_{1} α_{2} \dots α_{p}} \det [{(u_{μ})}^{λ}] . \end{aligned}

CHAPTER
5 STRESS-ENERGY TENSOR AND CONSERVATION LAWS

§5.1. TRACK-1 OVERVIEW

"Geometry tells matter how to move, and matter tells geometry how to curve." However, it will do no good to look into curvature (Part III) and Einstein's law for the production of curvature by mass-energy (Part IV) until a tool can be found to determine how much mass-energy there is in a unit volume. That tool is the stress-energy tensor. It is the focus of attention in this chapter.
The essential features of the stress-energy tensor are summarized in Box 5.1 for the benefit of readers who want to rush on into gravitation physics as quickly as
possible. Such readers can proceed directly from Box 5.1 into Chapter 6-though by doing so, they close the door on several later portions of track two, which lean
heavily on material treated in this chapter.

§5.2. THREE-DIMENSIONAL VOLUMES AND DEFINITION OF THE STRESS-ENERGY TENSOR

The rest of this chapter is

Track 2.

It depends on no preceding

Track-2 material.

It is needed as preparation for Chapter 20 (conservation laws for mass and angular momentum).

It will be extremely helpful in all applications of gravitation theory (Chapters 18-40).

Spacetime contains a flowing "river" of 4-momentum. Each particle carries its 4-momentum vector with itself along its world line. Many particles, on many world lines, viewed in a smeared-out manner (continuum approximation), produce a continuum flow-a river of 4-momentum. Electromagnetic fields, neutrino fields, meson fields: they too contribute to the river.

How can the flow of the river be quantified? By means of a linear machine: the stress-energy tensor

T

Choose a small, three-dimensional parallelepiped in spacetime with vectors

A

B, C

for edges (Figure 5.1). Ask how much 4 -momentum crosses that volume in

Box 5.1 CHAPTER 5 SUMMARIZED

A. STRESS-ENERGY TENSOR AS A MACHINE

At each event in spacetime, there exists a stress-energy tensor. It is a machine that contains a knowledge of the energy density, momentum density, and stress as measured by any and all observers at that event. Included are energy, momentum, and stress associated with all forms of matter and all nongravitational fields.
The stress-energy tensor is a linear, symmetric machine with two slots for the insertion of two vectors:

T (\dots, \dots)

. Its output, for given input, can be summarized as follows.
(1) Insert the 4-velocity

u

of an observer into one of the slots; leave the other slot empty. The output is

T (u, \dots) = T (\dots, u) = - (\begin{array}{l} density of 4-momentum, \\ "d d p / d V, " i.e., 4-momentum \\ per unit of three-dimensional volume, \\ as measured in observer's \\ Lorentz frame at event where \\ T is chosen \end{array});

i.e.,

T^{α}_{β} u^{β} = T_{β}^{α} u^{β} = - (d p^{α} / d V)

for observer with 4 -velocity

u^{α}

.
(2) Insert 4-velocity of observer into one slot; insert an arbitrary unit vector

n

into the other slot. The output is

T (u, n) = T (n, u) = - (\begin{array}{l} component, " n \cdot d p / d V ", of \\ 4 -momentum density along the \\ n direction, as measured in \\ observer's Lorentz frame \end{array})

i.e.,

T_{α β} u^{α} n^{β} = T_{α β} n^{α} u^{β} = - n_{μ} d p^{μ} / d V

.
(3) Insert 4-velocity of observer into both slots. The output is the density of massenergy that he measures in his Lorentz frame:

T (u, u) = (\binom{mass-energy per unit volume as measured}{in frame with 4 -velocity u})

(4) Pick an observer and choose two spacelike basis vectors,

e_{j}

and

e_{k}

, of his Lorentz frame. Insert

e_{j}

and

e_{k}

into the slots of

T

. The output is the

j, k

component of the stress as measured by that observer:

T_{j k} = T (e_{j}, e_{k}) = T_{k j} = T (e_{k}, e_{j})

= (\begin{array}{l} j -component of force acting \\ from side x^{k} - ε to side x^{k} + ε, \\ across a unit surface area with \\ perpendicular direction e_{k} \end{array}) = (\begin{array}{l} k -component of force acting \\ from side x^{j} - ε to side x^{j} + ε, \\ across a unit surface area with \\ perpendicular direction e_{j} \end{array}) .

Box 5.1 (continued)

B. STRESS-ENERGY TENSOR FOR A PERFECT FLUID

One type of matter studied extensively later in this book is a "perfect fluid." A perfect fluid is a fluid or gas that (1) moves through spacetime with a 4 -velocity

u

which may vary from event to event, and (2) exhibits a density of mass-energy

ρ

and an isotropic pressure

p

in the rest frame of each fluid element. Shear stresses, anisotropic pressures, and viscosity must be absent, or the fluid is not perfect. The stress-energy tensor for a perfect fluid at a given event can be constructed from the metric tensor,

g

, the 4 -velocity,

u

, and the rest-frame density and pressure,

ρ

and

p

T = (ρ + p) u \otimes u + p g, or T_{α β} = (ρ + p) u_{α} u_{β} + p g_{α β}

In the fluid's rest frame, the components of this stress-energy tensor have the expected form (insert into a slot of

T

, as 4-velocity of observer, just the fluid's 4 -velocity):

T^{α}_{β} u^{β} = [(ρ + p) u^{α} u_{β} + p δ^{α}_{β}] u^{β} = - (ρ + p) u^{α} + p u^{α} = - ρ u^{α};

i.e.,

\begin{aligned} T^{0}_{β} u^{β} = - ρ = - (mass-energy density) = - d p^{0} / d V, \\ T_{_{β}}^{j} u^{β} = 0 = - (momentum density) = - d p^{j} / d V \end{aligned}

also

T_{j k} = T (e_{j}, e_{k}) = p δ_{j k} = stress-tensor components.

C. CONSERVATION OF ENERGY-MOMENTUM

In electrodynamics the conservation of charge can be expressed by the differential equation

\partial (charge density) / \partial t + \nabla \cdot (current density) = 0

i.e.,

J^{0}_{, 0} + \nabla \cdot J = 0

; i.e.

J^{α}_{, α} = 0

; i.e.,

\nabla \cdot J = 0

. Similarly, conservation of energy-momentum can be expressed by the fundamental geometric law

\nabla \cdot T = 0

(Because

T

is symmetric, it does not matter on which slot the divergence is taken.) This law plays an important role in gravitation theory.

Figure 5.1.

The "river" of 4 -momentum flowing through spacetime, and three different 3 -volumes across which it flows. (One dimension is suppressed from the picture; so the 3 -volumes look like 2 -volumes.) The first 3 -volume is the interior of a cubical soap box momentarily at rest in the depicted Lorentz frame. Its edges are

L e_{x}, L e_{y}, L e_{z}

; and its volume 1 -form, with "positive" sense toward future ("standard orientation"), is

Σ = L^{3} d t = - V u (V = L^{3} =

volume as measured in rest frame;

u = - d t = 4

-velocity of box). The second 3 -volume is the "world sheet" swept out in time

Δ τ

by the top of a second cubical box. The box top's edges are

L e_{x}

and

L e_{z}

; and its volume 1 -form, with "positive" sense away from the box's interior, in direction of increasing

y

, is

Σ = L^{2} Δ τ d y = a Δ τ σ

(

a = L^{2} =

area of box top;

σ = d y =

unit 1 -form containing world tube). The third 3 -volume is an arbitrary one, with edges

A

B, C

and volume 1-form

Σ_{μ} = ϵ_{μ α β γ} A^{α} B^{β} C^{γ}

.
its positive sense (i.e., from its "negative side" toward its "positive side"). To calculate the answer: (1) Construct the "volume 1-form"

\begin{matrix} (5.1) & Σ_{μ} = + ϵ_{μ α β γ} A^{α} B^{β} C^{γ} \end{matrix}

the parallelepiped lies in one of the 1 -form surfaces, and the positive sense across the parallelepiped is defined to be the positive sense of the 1 -form

Σ

. (2) Insert this volume 1 -form into the second slot of the stress-energy tensor

T

. The result is

\begin{aligned} (5.2) & T (\dots, Σ) = p = (\binom{momentum crossing from}{negative side toward positive side}) . \\ empty ↑ \\ slot \end{aligned}

(3) To get the projection of the 4-momentum along a vector

w

or 1 -form

α

, insert the volume 1-form

Σ

into the second slot and

w

α

into the first:

\begin{matrix} (5.3) & T (w, Σ) = w \cdot p, T (α, Σ) = ⟨ α, p ⟩ \end{matrix}

This defines the stress-energy tensor.

Mathematical representation of 3 -volumes

Momentum crossing a 3 -volume calculated, using stress-energy tensor

The key features of 3 -volumes and the stress-energy tensor are encapsulated by the above three-step procedure. But encapsulation is not sufficient; deep understanding is also required. To gain it, one must study special cases, both of 3 -volumes and of the operation of the stress-energy machinery.

A Special Case

Interior of a soap box:

Its volume 1-form

Its 4-momentum content

Its energy density

A soap box moves through spacetime. A man at an event

P_{0}

on the box's world line peers inside it, and examines all the soap, air, and electromagnetic fields it contains. He adds up all their 4-momenta to get a grand total

p_{boxat 9_{0}}

. How much is this grand total? One can calculate it by noting that the 4 -momentum inside the box at

P_{0}

is precisely the 4 -momentum crossing the box from past toward future there (Figure 5.1). Hence, the 4 -momentum the man measures is

\begin{matrix} (5.4) & p_{box at Q_{0}} = T (\dots, Σ) \end{matrix}

where

Σ

is the box's volume 1 -form at

P_{0}

. But for such a soap box,

Σ

has a magnitude equal to the box's volume

V

as measured by a man in its momentary rest frame, and the box itself lies in one of the hyperplanes of

Σ

; equivalently,

\begin{matrix} (5.5) & Σ = - V u \end{matrix}

where

u

is the soap box's 4 -velocity at

P_{0}

(minus sign because

u

, regarded as a 1 -form, has positive sense toward the past,

u_{0} < 0

); see Box 5.2. Hence, the total 4 -momentum inside the box is

\begin{matrix} (5.6) & p_{boxat φ_{0}} = T (\dots, - V u) = - V T (\dots, u) \end{matrix}

or, in component notation,

\begin{matrix} (5.6') & {(p^{α})}_{box at g_{0}} = - V T^{α β} u_{β} \end{matrix}

The energy in the box, as measured in its rest frame, is minus the projection of the 4 -momentum on the box's 4 -velocity:

E = - u \cdot p_{boxat P_{0}} = + V T^{α β} u_{α} u_{β} = V T (u, u)

\begin{aligned} (5.7) & energy density as \\ (\binom{measured in box's}{rest frame}) = \frac{E}{V} = T (u, u) . \end{aligned}

Another Special Case

A man riding with the same soap box opens its top and pours out some soap. In a very small interval of time

Δ τ

, how much total 4-momentum flows out of the box?

Box 5.2 THREE-DIMENSIONAL VOLUMES

A. General Parallelepiped

Edges of parallelepiped are three vectors $A$ , $B, C$ . One must order the edges; e.g., " $A$ is followed by $B$ is followed by $C$ ."

(One dimension, that orthogonal to the parallelepiped, is suppressed here.)
Volume trivector is defined to be $A \land B \land C$ . It enters into the sophisticated theory of volumes (Chapter 4), but is not used much in the elementary theory.
Volume 1-form is defined by $Σ_{μ} = ϵ_{μ α β γ} A^{α} B^{β} C^{γ}$ . ( $A, B, C$ must appear here in standard order as chosen in step 1.) Note that the vector "corresponding" to $Σ$ and the volume trivector are related by $Σ = -^{*} (A \land B \land C)$ .
Orientation of the volume is defined to agree with the orientation of its 1 -form $Σ$ . More specifically: the edges $A, B, C$ lie in a hyperplane of $Σ (⟨ Σ, A ⟩ = ⟨ Σ, B ⟩ = ⟨ Σ, C ⟩ = 0$ ; no "bongs of bell"). Thus, the volume itself is one of $Σ$ 's hyperplanes! The positive sense moving away from the volume is defined to be the positive sense of $Σ$ . Note: reversing the order of $A, B, C$ reverses the positive

(One dimension, that along which $C$ extends, is suppressed here.) sense!
The "standard orientation" for a spacelike 3-volume has the positive sense of the 1-form $Σ$ toward the future, corresponding to $A, B, C$ forming a righthanded triad of vectors.
B. 3-Volumes of Arbitrary Shape

Can be analyzed by being broken up into union of parallelepipeds.

C. Interior of a Soap Box (Example)

Analysis in soap box's rest frame. Pick an event on the box's world line. The box's three edges there are three specific vectors $A, B, C$ . In the box's rest frame they are purely spatial: $A^{0} = B^{0} = C^{0} = 0$ . Hence, the volume 1-form has components $Σ_{j} = 0$ and
$Σ_{0} = ϵ_{0 i j k} A^{i} B^{j} C^{k} = \det ‖ \begin{array}{lll} A^{1} & A^{2} & A^{3} \\ B^{1} & B^{2} & B^{3} \\ C^{1} & C^{2} & C^{3} \end{array} ‖$
$= A \cdot (B \times C)$ , in the standard notation of 3-dimensional vector analysis;
$= + V (V =$ volume of box) if $(A, B, C)$ are righthand ordered (positive sense of $Σ$ toward future; standard orientation);
$= - V (V =$ volume of box $)$ if $(A, B, C)$ are lefthand ordered (positive sense of $Σ$ toward past).
This result reexpressed in geometric language: Let $u$ be the box's 4-velocity and $V$ be its volume, as measured in its rest frame. Then either
$Σ = - V u$ , in which case the "positive side" of the box's 3 -surface is the future side, and its edges are ordered in a righthanded manner-the standard orientation;
or else
$Σ = + V u$ , in which case the "positive side" is the past side, and the box's edges are ordered in a lefthanded manner.

D. 3-Volume Swept Out in Time $Δ τ$ by Two-Dimensional Top of a Soap Box (Example)

Analysis in box's rest frame: Pick an event on box's world line. There the two edges of the box top are vectors $A$ and $B$ . In the box's rest frame, orient the space axes so that $A$ and $B$ lie in the $y, z$ -plane. During the lapse of a proper time $Δ τ$ , the box top sweeps out
a 3 -volume whose third edge is $u Δ τ (u =$ 4 -velocity of box). In the box's rest-frame, with ordering " $A$ followed by $B$ followed by $u Δ τ$ , " the volume 1 -form has components
$Σ_{0} = Σ_{2} = Σ_{3} = 0$ , and
$Σ_{1} = ϵ_{1 j k 0} A^{j} B^{k} Δ τ u^{0} = - ϵ_{01 j k} A^{j} B^{k} Δ τ$
$= - a Δ τ (a =$ area of box top $)$ if $(e_{x}, A, B)$ are righthand ordered
$= + a Δ τ (a =$ area of box top $)$ if $(e_{x}, A, B)$ are lefthand ordered.
(Note: No standard orientation can be defined in this case, because $Σ$ can be carried continuously into $- Σ$ by purely spatial rotations.)
This result reexpressed in geometric language: Let $a$ be the area of the box top as measured in its rest frame; and let $σ$ be a unit 1-form, one of whose surfaces contains the box top and its 4 -velocity (i.e., contains the box top's "world sheet"). Orient the positive sense of $σ$ with the (arbitrarily chosen) positive sense of the box-top 3 -volume. Then

Σ = a Δ τ σ

To answer this question, consider the three-dimensional volume swept out during

Δ τ

by the box's opened two-dimensional top ("world sheet of top"). The 4-momentum asked for is the 4 -momentum that crosses this world sheet in the positive sense (see Figure 5.1); hence, it is

\begin{matrix} (5.8) & p_{flows out} = T (\dots, Σ), \end{matrix}

where

Σ

is the world sheet's volume 1 -form. Let

a

be the area of the box top, and

σ

be the outward-oriented unit 1-form, whose surfaces contain the world sheet (i.e., contain the box top and its momentary 4 -velocity vector). Then

\begin{matrix} (5.9) & Σ = a Δ τ σ \end{matrix}

(see Box 5.2 ); so the 4 -momentum that flows out during

Δ τ

\begin{matrix} (5.10) & p_{flows out} = a Δ τ T (\dots, σ) \end{matrix}

§5.3. COMPONENTS OF STRESS-ENERGY TENSOR

Like all other tensors, the stress-energy tensor is a machine whose definition and significance transcend coordinate systems and reference frames. But any one observer, locked as he is into some one Lorentz frame, pays more attention to the components of

T

than to

T

itself. To each component he ascribes a specific physical significance. Of greatest interest, perhaps, is the "time-time" component. It is the total density of mass-energy as measured in the observer's Lorentz frame:

\begin{matrix} (5.11) & T_{00} = - T_{0}^{0} = T^{00} = T (e_{0}, e_{0}) = density of mass-energy \end{matrix}

(cf. equation 5.7, with the observer's 4-velocity

u

replaced by the basis vector

e_{0} = u

).
The "spacetime" components

T^{j 0}

can be interpreted by considering the interior of a soap box at rest in the observer's frame. If its volume is

V

, then its volume 1 -form is

Σ = - V u = + V d t

; and the

μ

-component of 4 -momentum inside it is

p^{μ} = ⟨ d x^{μ}, p ⟩ = T (d x^{μ}, Σ) = V T (d x^{μ}, d t) = V T^{μ 0} .

Thus, the 4 -momentum per unit volume is

\begin{matrix} (5.12a) & p^{μ} / V = T^{μ 0}, \end{matrix}

or, equivalently:

\begin{aligned} (5.13a) & T^{00} = & density of mass-energy \\ (units: g / {cm}^{3}, or erg / {cm}^{3}, or {cm}^{- 2}); \\ (5.13b) & T^{j 0} = & density of j -component of momentum \\ (units: g (cm / \sec) {cm}^{- 3}, or {cm}^{- 2}) \end{aligned}

The top of a soap box:

Its volume 1-form

Its 4-momentum that flows across

Physical interpretation of stress-energy tensor's components:

T^{0 k}

: energy flux

T^{j k}

: stress

Number-flux vector for swarm of particles defined

During the lapse of time

Δ t

, this 2 -surface sweeps out a 3 -volume with volume 1 -form

Σ = a Δ t d x^{k}

(see Box 5.2). The

μ

-component of 4 -momentum that crosses the 2 -surface in time

Δ t

p^{μ} = T (d x^{μ}, Σ) = a Δ t T (d x^{μ}, d x^{k}) = a Δ t T^{μ k}

Thus, the flux of 4-momentum (4-momentum crossing a unit surface oriented perpendicular to

e_{k}

, in unit time) is

\begin{matrix} (5.12b) & {(p^{μ} / a Δ t)}_{crossing surface ⊥ to e_{k}} = T^{μ k} \end{matrix}

or, equivalently:

\begin{aligned} T^{0 k} = & k -component of energy flux \\ (units: erg / {cm}^{2} \sec, or {cm}^{- 2}); \\ T^{j k} = & j, k component of "stress" \\ \equiv & k -component of flux of j -component of momentum \\ \equiv & j -component of force produced by fields and matter at x^{k} - ϵ acting \\ on fields and matter at x^{k} + ϵ across a unit surface, the perpendicular \\ to which is e_{k} \\ (units: dynes / {cm}^{2}, or {cm}^{- 2}). \end{aligned}

(Recall that "momentum transfer per second" is the same as "force.")
The stress-energy tensor is necessarily symmetric,

T^{α β} = T^{β α}

; but the proof of this will be delayed until several illustrations have been examined.

§5.4. STRESS-ENERGY TENSOR FOR A SWARM OF PARTICLES

Consider a swarm of particles. Choose some event

P

inside the swarm. Divide the particles near

P

into categories,

A = 1, 2, \dots

, in such a way that all particles in the same category have the same properties:

$m_{(A)}$ ,	rest mass;
$u_{(A)}$ ,	4-velocity;
$p_{(A)} = m_{(A)} u_{(A)}$ ,	4-momentum.

Let

N_{(A)}

be the number of category-

A

particles per unit volume, as measured in the particles' own rest frame. Then the "number-flux vector"

S_{(A)}

, defined by

\begin{matrix} (5.14) & s_{(A)} \equiv N_{(A)} u_{(A)} \end{matrix}

has components with simple physical meanings. In a frame where category-

A

particles have ordinary velocity

v_{(A)}

, the meanings are:

\begin{matrix} (5.15b) & S_{(A)} = N_{(A)} u_{(A)} = S_{(A)}^{0} v_{(A)} = flux of particles. \end{matrix}

Consequently, the 4-momentum density has components

\begin{aligned} T_{(A)}^{μ 0} & = p_{(A)}^{μ} S_{(A)}^{0} = m_{(A)} u_{(A)}^{μ} N_{(A)} u_{(A)}^{0} \\ = m_{(A)} N_{(A)} u_{(A)}^{μ} u_{(A)}^{0} \end{aligned}

and the flux of

μ

-component of momentum across a surface with perpendicular direction

e_{j}

\begin{aligned} T_{(A)}^{μ j} & = p_{(A)}^{μ} S_{(A)}^{j} = m_{(A)} u_{(A)}^{μ} N_{(A)} u_{(A)}^{j} \\ = m_{(A)} N_{(A)} u_{(A)}^{μ} u_{(A)}^{j} . \end{aligned}

These equations are precisely the

μ, 0

and

μ, j

components of the geometric, frameindependent equation

\begin{matrix} (5.16) & T_{(A)} = m_{(A)} N_{(A)} u_{(A)} \otimes u_{(A)} = p_{(A)} \otimes S_{(A)} \end{matrix}

The total number-flux vector and stress-energy tensor for all particles in the swarm near

P

are obtained by summing over all categories:

\begin{aligned} (5.17) & S & = \sum_{A} N_{(A)} u_{(A)} \\ (5.18) & T & = \sum_{A} m_{(A)} N_{(A)} u_{(A)} \otimes u_{(A)} = \sum_{A} p_{(A)} \otimes S_{(A)} \end{aligned}

§5.5. STRESS-ENERGY TENSOR FOR A PERFECT FLUID

There is no simpler example of a fluid than a gas of noninteracting particles ("ideal gas") in which the velocities of the particles are distributed isotropically. In the Lorentz frame where isotropy obtains, symmetry argues equality of the diagonal space-space components of the stress-energy tensor,

\begin{matrix} (5.19) & T_{x x} = T_{y y} = T_{z z} = \sum_{A} \frac{m_{(A)} v_{x (A)}}{{(1 - v_{(A)}^{2})}^{1 / 2}} \frac{N_{(A)} v_{x (A)}}{{(1 - v_{(A)})}^{2 / 2}} \end{matrix}

and vanishing of all the off-diagonal components. Moreover, (5.19) represents a product: the number of particles per unit volume, multiplied by velocity in the

x

-direction (giving flux in the

x

-direction) and by momentum in the

x

-direction,

Stress-energy tensor for swarm of particles

Ideal gas defined
giving the standard kinetic-theory expression for the pressure,

p

. Therefore, the stress-energy tensor takes the form

\begin{matrix} (5.20) & T_{α β} = ‖ \begin{array}{llll} ρ & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{array} ‖ \end{matrix}

in this special Lorentz frame-the "rest frame" of the gas. Here the quantity

ρ

has nothing directly to do with the rest-masses of the constituent particles. It measures the density of rest-plus-kinetic energy of these particles.

Rewrite (5.20) in terms of the 4-velocity

u^{α} = (1, 0, 0, 0)

of the fluid in the gas's rest frame, and find

\begin{aligned} T_{α β} & = ‖ \begin{array}{llll} ρ & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} ‖ + ‖ \begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{array} ‖ \\ = ρ u_{α} u_{β} + p (η_{α β} + u_{α} u_{β}), \end{aligned}

Stress-energy tensor for ideal gas or perfect fluid
or, in frame-independent, geometric language

\begin{matrix} (5.21) & T = p g + (ρ + p) u \otimes u \end{matrix}

Expression (5.21) has general application. It is exact for the "ideal gas" just considered. It is also exact for any fluid that is "perfect" in the sense that it is free of such transport processes as heat conduction and viscosity, and therefore (in the rest frame) free of shear stress (diagonal stress tensor; diagonal components identical, because if they were not identical, a rotation of the frame of reference would reveal presence of shear stress). However, for a general perfect fluid, density

ρ

of mass-energy as measured in the fluid's rest frame includes not only rest mass plus kinetic energy of particles, but also energy of compression, energy of nuclear binding, and all other sources of mass-energy [total density of mass-energy as it might be determined by an idealized experiment, such as that depicted in Figure 1.12, with the sample mass at the center of the sphere, and the test particle executing oscillations of small amplitude about that location, with

ω^{2} = (4 π / 3) ρ]

§5.6. ELECTROMAGNETIC STRESS-ENERGY

Faraday, with his picture of tensions along lines of force and pressures at right angles to them (Figure 5.2), won insight into new features of electromagnetism. In addition to the tension

E^{2} / 8 π

(or

B^{2} / 8 π

) along lines of force, and an equal pressure at right angles, one has the Poynting flux

(E \times B) / 4 π

and the Maxwell expression for the

Figure 5.2.
Faraday stresses at work. When the electromagnet is connected to an alternating current, the aluminum ring flies into the air.
energy density,

(E^{2} + B^{2}) / 8 π

. All these quantities find their places in the Maxwell stress-energy tensor, defined by

\begin{matrix} (5.22) & 4 π T^{μ ν} = F^{μ α} F_{α}^{ν} - \frac{1}{4} η^{μ ν} F_{α β} F^{α β} . \end{matrix}

Stress-energy tensor for electromagnetic field

Exercise 5.1.

EXERCISE

Show that expression (5.22), evaluated in a Lorentz coordinate frame, gives

\begin{aligned} T^{00} & = (E^{2} + B^{2}) / 8 π, T^{0 j} = T^{j 0} = (E \times B)^{j} / 4 π \\ (5.23) & T^{j k} & = \frac{1}{4 π} [- (E^{j} E^{k} + B^{j} B^{k}) + \frac{1}{2} (E^{2} + B^{2}) δ^{j k}] \end{aligned}

Show that the stress tensor does describe a tension

(E^{2} + B^{2}) / 8 π

along the field lines and a pressure

(E^{2} + B^{2}) / 8 π

perpendicular to the field lines, as stated in the text.

§5.7. SYMMETRY OF THE STRESS-ENERGY TENSOR

All the stress-energy tensors explored above were symmetric. That they could not have been otherwise one sees as follows.

Calculate in a specific Lorentz frame. Consider first the momentum density (components

T^{j 0}

) and the energy flux (components

T^{0 j}

). They must be equal because

Proof that stress-energy tensor is symmetric energy

=

mass (

" E = M c^{2} = M

"):

\begin{aligned} T^{0 j} & = (energy flux) \\ = (energy density) \times (mean velocity of energy flow)^{j} \\ (5.24) & = (mass density) \times (mean velocity of mass flow)^{j} \\ = (momentum density) = T^{j 0} . \end{aligned}

Only the stress tensor

T^{j k}

remains. For it, one uses the same standard argument as in Newtonian theory. Consider a very small cube, of side

L

, mass-energy

T^{00} L^{3}

,
and moment of inertia

\sim T^{00} L^{5}

. With the space coordinates centered at the cube, the expression for the

z

-component of torque exerted on the cube by its surroundings is

\begin{aligned} τ^{z} = \underset{(\begin{array}{c} -component \\ of force on \\ + x face \end{array})}{\underset{⏟}{(- T^{y x} L^{2})}} \underset{\begin{array}{c} lever \\ arm to \\ + x face \end{array})}{\underset{⏟}{(L / 2)}} + \underset{(\begin{array}{c} y -component \\ of force on \\ - x face \end{array})}{\underset{⏟}{(T^{y x} L^{2})}} \underset{\begin{array}{c} lever \\ arm to \\ - x face \end{array})}{\underset{⏟}{(- L / 2)}} \\ - \underset{(\begin{array}{c} x -component \\ of force on \\ + y face \end{array})}{\underset{⏟}{(- T^{x y} L^{2})}} \underset{\begin{array}{c} lever \\ arm to \\ + y face \end{array})}{\underset{⏟}{(L / 2)}} - \underset{(\begin{array}{c} x -component \\ of force on \\ - y face \end{array})}{\underset{⏟}{(T^{x y} L^{2})}} \underset{(\begin{array}{c} lever \\ arm to \\ - y face \end{array})}{\underset{⏟}{(- L / 2)}} \\ = (T^{x y} - T^{y x}) L^{3} . \end{aligned}

Since the torque decreases only as

L^{3}

with decreasing

L

, while the moment of inertia decreases as

L^{5}

, the torque will set an arbitrarily small cube into arbitrarily great angular acceleration-which is absurd. To avoid this, the stresses distribute themselves so the torque vanishes:

T^{y x} = T^{x y} .

Put differently, if the stresses were not so distributed, the resultant infinite angular accelerations would instantaneously redistribute them back to equilibrium. This condition of torque balance, repeated for all other pairs of directions, is equivalent to symmetry of the stresses:

\begin{matrix} (5.25) & T^{j k} = T^{k j} . \end{matrix}

§5.8. CONSERVATION OF 4-MOMENTUM: INTEGRAL FORMULATION

Energy-momentum conservation has been a cornerstone of physics for more than a century. Nowhere does its essence shine forth so clearly as in Einstein's geometric formulation of it (Figure 5.3,a). There one examines a four-dimensional region of spacetime

V

bounded by a closed, three-dimensional surface

\partial V

. As particles and fields flow into

V

and later out, they carry 4-momentum. Inside

V

the particles collide, break up, radiate; radiation propagates, jiggles particles, produces pairs. But at each stage in this complex maze of physical processes, total energy-momentum is conserved. The energy-momentum lost by particles goes into fields; the energymomentum lost by fields goes into particles. So finally, when the "river" of 4-momentum exits from

V

, it carries out precisely the same energy-momentum as it carried in.
Restate this equality by asking for the total flux of 4-momentum outward across

\partial V

. Count inflowing 4 -momentum negatively. Then "inflow equals outflow" means "total outflow vanishes":

Integral conservation law for 4-momentum:

∮_{\partial r \cdot} T \cdot d^{3} Σ = 0

(a)

\partial \cdot V = S_{2} - S_{1}

(b)

(d)

\partial V = \partial (V_{2} - V_{1}) = \bar{S} - S

(c)

\partial V = \bar{S} - S + T

(e)
Figure 5.3.
(a) A four-dimensional region of spacetime

V

bounded by a closed three-dimensional surface

\partial V

. The positive sense of

\partial V

is defined to be everywhere outward (away from

V

). Conservation of energymomentum demands that every bit of 4 -momentum which flows into

V

through

\partial V

must somewhere flow back out; none can get lost inside; the interior contains no "sinks." Equivalently, the total flux of 4-momentum across

\partial V

in the positive (outward) sense must be zero:

∮_{\partial γ} T^{μ α} d^{3} Σ_{α} = 0 .

Figures (b), (c), (d), and (e) depict examples to which the text applies this law of conservation of 4 -momentum. All symbols

V

(or

S

) in these figures mean spacetime volumes (or spacelike 3 -volumes) with standard orientations. The dotted arrows indicate the positive sense of the closed surface

\partial V

used in the text's discussion of 4-momentum conservation. How

\partial V

is constructed from the surfaces

S

and

T

is indicated by formulas below the figures. For example, in case (b),

\partial V = S_{2} - S_{1}

means that

\partial V

is made by joining together

S_{2}

with its standard orientation and

S_{1}

with reversed orientation.

Total flux of 4-momentum outward across a closed three-dimensional surface must vanish.

To calculate the total outward flux in the most elementary of fashions, approximate the closed 3 -surface

\partial V

by a large number of flat 3 -volumes ("boiler plates") with positive direction oriented outward (away from

V

). Then

\begin{matrix} (5.27) & p_{total out} = \sum_{boiler plates A} T (\dots, Σ_{(A)}) = 0, \end{matrix}

where

Σ_{(1)}

is the volume 1 -form of boiler plate

A

. Equivalently, in component notation

\begin{matrix} (5.27') & p_{total out}^{μ} = \sum_{A} T^{μ α} Σ_{(A) α} \end{matrix}

To be slightly more sophisticated about the calculation, take the limit as the number of boiler plates goes to infinity and their sizes go to zero. The result is an integral (Box 5.3, at the end of this section),

\begin{matrix} (5.28) & p_{totalout}^{μ} = ∮_{\partial V} T^{μ α} d^{3} Σ_{α} = 0 \end{matrix}

Think of this (like all component equations) as a convenient way to express a coordinate-independent statement:

\begin{matrix} (5.29) & p_{total out} = ∮_{\partial V r} T \cdot d^{3} Σ = 0 \end{matrix}

To be more sophisticated yet (not recommended on first reading of this book) and to simplify the computations in practical cases, interpret the integrands as exterior differential forms (Box 5.4, at the end of this section).

But however one calculates it, and however one interprets the integrands, the statement of the result is simple: the total flux of 4 -momentum outward across a closed 3 -surface must vanish.
Several special cases of this "integral conservation law," shown in Figure 5.3, are instructive. There shown, in addition to the general case (a), are:

Case (b)

The closed 3-surface

\partial V

is made up of two slices taken at constant time

t

of a specific Lorentz frame, plus timelike surfaces at "infinity" that join the two slices together. The surfaces at infinity do not contribute to

∮_{\partial V} T^{μ α} d^{3} Σ_{α}

if the stress-energy tensor dies out rapidly enough there. The boundary

\partial V

of the standard-oriented 4 -volume

V

, by definition, has its positive sense away from

V

. This demands nonstandard
orientation of

S_{1}

(positive sense toward past), as is indicated by writing

\partial V =

S_{2} - S_{1}

; and it produces a sign flip in the evaluation of the hypersurface integral

0 = ∮_{\partial V} T^{α μ} d^{3} Σ_{μ} = - \int_{S_{1}} T^{α 0} d x d y d z + \int_{S_{2}} T^{α 0} d x d y d z

Because

T^{α 0}

is the density of 4-momentum, this equation says

\begin{aligned} (5.30) & (\binom{total 4-momentum in}{all of space at time t_{1}}) & = \int_{S_{1}} T^{α 0} d x d y d z \\ = (\binom{total 4-momentum in}{all of space at time t_{2}}) = \int_{S_{2}} T^{α 0} d x d y d z \end{aligned}

Case (c)

Here one wants to compare hypersurface integrals over

S

and

\overset{―}{S}

, which are slices of constant time,

t =

const and

\bar{t} =

const in two different Lorentz frames. To form a closed surface, one adds time-like hypersurfaces at infinity and assumes they do not contribute to the integral. The orientations fit together smoothly and give a closed surface

\partial V = \bar{S} - S + (surfaces at infinity)

only if one takes

V = V_{2} - V_{1}

-i.e., only if one uses the nonstandard 4-volume orientation in

V_{1}

. (See part A. 1 of Box 5.3 for "standard" versus "non-standard" orientation.) The integral conservation law then gives

0 = \int_{\bar{S}} T \cdot d^{3} Σ - \int_{S} T \cdot d^{3} Σ

or, equivalently,

\begin{aligned} (5.31) & \int_{\bar{S}} T \cdot & d^{3} Σ = (total 4 -momentum p on \bar{S}) \\ = \int_{S} T \cdot d^{3} Σ = (total 4-momentum p on S) \end{aligned}

This says that observers in different Lorentz frames measure the same total 4-momentum

p

. It does not mean that they measure the same components

(p^{α} \neq p^{\bar{α}})

; rather, it means they measure the same geometric vector

p_{on S} = p^{α} e_{α} = p_{on \bar{S}} = p^{\bar{α}} e_{\bar{α}}

a vector whose components are connected by the usual Lorentz transformation law

\begin{matrix} (5.32) & p^{α} = Λ^{α}_{\bar{β}} p^{\bar{β}} \end{matrix}

Total 4-momentum the same in all Lorentz frames

Total 4-momentum independent of hypersurface where measured

Change with time of 4-momentum in a box equals flux of 4 -momentum across its faces

Differential conservation law for 4-momentum:

\nabla \cdot T = 0

Case (d)

Here the contribution to the integral comes entirely from two arbitrary spacelike hypersurfaces,

S_{A}

and

S_{B}

, cutting all the way across spacetime. As in cases (a) and (b), the integral form of the conservation law says

\begin{matrix} (5.33) & p_{on S_{A}} = p_{on S_{B}}; \end{matrix}

i.e., the total 4-momentum on a spacelike slice through spacetime is independent of the specific slice chosen-so long as the energy-momentum flux across the "hypersurface at infinity" connecting

S_{A}

and

S_{B}

is zero.

Case (e)

This case concerns a box whose walls oscillate and accelerate as time passes. The three-dimensional boundary

\partial V

is made up of (1) the interior

S

of the box, at an initial moment of time

t =

constant in the box's initial Lorentz frame, taken with nonstandard orientation; (2) the interior

\bar{S}

of the box, at

\bar{t} =

constant in its final Lorentz frame, with standard orientation; (3) the 3-volume

T

swept out by the box's two-dimensional faces between the initial and final states, with positive sense oriented outward. The integral conservation law

\int_{\partial V} T \cdot d^{3} Σ = 0

says

(\binom{total 4-momentum}{in box at \bar{S}}) - (\binom{total 4-momentum}{in box at S})

\begin{matrix} (5.34) & = (\binom{total 4 -momentum that enters box through}{its faces between states S and \bar{S}}) . \end{matrix}

§5.9. CONSERVATION OF 4-MOMENTUM: DIFFERENTIAL FORMULATION

Complementary to any "integral conservation law in flat spacetime" is a "differential conservation law" with identical information content. To pass back and forth between them, one can use Gauss's theorem.

Gauss's theorem in four dimensions, applied to the law of 4-momentum conservation, converts the surface integral of

T^{μ α}

into a volume integral of

T^{μ α}_{, α}

\begin{matrix} (5.35) & 0 = ∮_{\partial V} T^{μ α} d^{3} Σ_{α} = \int_{V} T_{, α}^{μ α} d t d x d y d z \end{matrix}

(See Box 5.3 for elementary discussion; Box 5.4 for sophisticated discussion.) If the integral of

T^{μ α}_{, α}

is to vanish, as demanded, for any and every 4 -volume

V

, then

T^{μ α}_{, α}

must itself vanish everywhere in spacetime:

\begin{matrix} (5.36) & T_{, α}^{μ α} = 0; i.e., \nabla \cdot T = 0 everywhere. \end{matrix}

Box 5.3 VOLUME INTEGRALS, SURFACE INTEGRALS, AND GAUSS'S THEOREM IN COMPONENT NOTATION

A. Volume Integrals in Spacetime

By analogy with three-dimensional space, the volume of a "hyperparallelepiped" with vector edges $A, B, C, D$ is

\begin{aligned} 4-volume & \equiv Ω \equiv ϵ_{α β γ δ} A^{α} B^{β} C^{γ} D^{δ} = \det ‖ \begin{array}{llll} A^{0} & A^{1} & A^{2} & A^{3} \\ B^{0} & B^{1} & B^{2} & B^{3} \\ C^{0} & C^{1} & C^{2} & C^{3} \\ D^{0} & D^{1} & D^{2} & D^{3} \end{array} ‖ \\ =^{*} (A \land B \land C \land D) . \end{aligned}

Here, as for 3-volumes, orientation matters; interchange of any two edges reverses the sign of

Ω

. The standard orientation for any 4 -volume is the one which makes

Ω

positive; thus,

e_{0} \land e_{1} \land e_{2} \land e_{3}

has standard orientation if

e_{0}

points toward the future and

e_{1}, e_{2}, e_{3}

are a righthanded triad.
2. The "volume element" whose edges in a specific, standard-oriented Lorentz frame are

A^{α} = (Δ t, 0, 0, 0), B^{α} = (0, Δ x, 0, 0), C^{α} = (0, 0, Δ y, 0), D^{α} = (0, 0, 0, Δ z)

has a 4 -volume, according to the above definition, given by

Δ^{4} Ω = ϵ_{0123} Δ t Δ x Δ y Δ z = Δ t Δ x Δ y Δ z .

Thus, the volume integral of a tensor $S$ over a four-dimensional region $V$ of spacetime, defined as

M \equiv \underset{\begin{matrix} Lim \end{matrix} \underset{(\begin{matrix} number of \\ elementary \\ volumes \end{matrix}}{\to \infty} \underset{(\begin{matrix} elementary \\ volumes C \\ in V \end{matrix})}{} S_{at center of C} (volume of a),}{}

can be calculated in a Lorentz frame by

M^{α}_{β γ} = \int_{γ} S^{α}_{β γ} d^{4} Ω = \int_{γ} S^{α}_{β γ} d t d x d y d z

Box 5.3 (continued)

B. Integrals over 3-Surfaces in Spacetime

Introduce arbitrary coordinates $a, b, c$ on the three-dimensional surface. The elementary volume bounded by coordinate surfaces
$a_{0} < a < a_{0} + Δ a, b_{0} < b < b_{0} + Δ b$ , $c_{0} < c < c_{0} + Δ c$
has edges

A^{α} = \frac{\partial x^{α}}{\partial a} Δ a, B^{β} = \frac{\partial x^{β}}{\partial b} Δ b, C^{γ} = \frac{\partial x^{γ}}{\partial c} Δ c;

so its volume 1-form is

Δ^{3} Σ_{μ} = ϵ_{μ α β γ} \frac{\partial x^{α}}{\partial a} \frac{\partial x^{β}}{\partial b} \frac{\partial x^{γ}}{\partial c} Δ a Δ b Δ c .

The integral of a tensor $S$ over the 3 -surface $S$ thus has components

N^{α}_{β} = \int_{S} S^{α}_{β} γ^{3} Σ_{γ} = \int_{S} S^{α}_{β} γ_{ϵ_{γ μ ν λ}} \frac{\partial x^{μ}}{\partial a} \frac{\partial x^{ν}}{\partial b} \frac{\partial x^{λ}}{\partial c} d a d b d c .

An equivalent formula involving a Jacobian is often used (see exercise 5.5):

N^{α}_{β} = \int_{S} S^{α}_{β}^{γ} \frac{1}{3!} ϵ_{γ μ ν λ} \frac{\partial (x^{μ}, x^{ν}, x^{λ})}{\partial (a, b, c)} d a d b d c .

C. Gauss's Theorem Stated

Consider a bounded four-dimensional region of spacetime $V$ with closed boundary $\partial V$ . Orient the volume 1-forms on $\partial V$ so that the "positive sense" is away from $V$ .
Choose a tensor field $S$ . Integrate its divergence over $V$ , and integrate it itself over $\partial V$ . The results must be the same (Gauss's theorem):

\int_{V} S^{α}_{β}^{γ}, γ^{4} d^{4} Ω = ∮_{\partial V} S^{α}_{β}^{γ} d^{3} Σ_{γ} .

D. Proof of Gauss's Theorem

The indices $α$ and $β$ of $S^{α}_{β} γ$ "go along for a free ride," so one can suppress them from the proof. Then the equation to be derived is

\int_{\tilde{γ}} S^{γ}_{, γ} d t d x d y d z = ∮_{\partial γ} S^{γ} d^{3} Σ_{γ} .

Since the integral of a derivative is just the original function, the volume integral of $S^{0}_{0}$ is

\begin{aligned} \int_{V} S^{0}_{, 0} d t d x d y d z \\ = \int_{"up"} S^{0} d x d y d z - \int_{("down"} S^{0} d x d y d z \end{aligned}

3. The surface integral

\int_{\partial τ} S^{0} d^{3} Σ_{0}

can be reduced to the same set of terms:
a. Use

x, y, z

as coordinates on

\partial V

. On the "up" side,

d^{3} Σ_{0}

must be positive to achieve a "positive" sense pointing away from

V

, so (see part B above)

d^{3} Σ_{0} = ϵ_{0 α β γ} \frac{\partial x^{α}}{\partial x} \frac{\partial x^{β}}{\partial y} \frac{\partial x^{γ}}{\partial z} d x d y d z = ϵ_{0123} d x d y d z = d x d y d z

b. On the "down" side,

d^{3} Σ_{0}

must be negative, so

d^{3} Σ_{0} = - d x d y d z

c. Hence,

\int_{\partial r} S^{0} d^{3} Σ_{0} = \int_{"up"} S^{0} d x d y d z - \int_{"down"} S^{0} d x d y d z

Equality is proved for the other components in the same manner. Adding components produces the result desired:

\int_{γ} S^{γ}_{, γ} d^{4} Ω = ∮_{\partial γ} S^{γ} d^{3} Σ_{γ}

FOR THE READER WHO HAS STUDIED CHAPTER 4

Box 5.4 I. EVERY INTEGRAL IS THE INTEGRAL OF A FORM.
II. THE THEOREM OF GAUSS IN THE LANGUAGE OF FORMS.
I. Every integral encountered in Chapter 5 can be interpreted as the integral of an exterior differential form. This circumstance shows up in fourtold and threctold integrals, for example, in the fact that

d^{4} Ω = ε =^{*} 1 = ε_{0123} d t \land d x \land d y \land d z

and

d^{3} Σ_{μ} = ε_{μ | α β γ |} d x^{α} \land d x^{β} \land d x^{γ}

are basis 4- and 3-forms. (Recall: the indices

α β γ

between vertical bars are to be summed only over

0 \leq α < β < γ \leq 3

.) A more extensive glossary of notations is found in C below.
II. Gauss's Theorem for a tensor integral in flat space reads

\int_{V} (\nabla \cdot S) d^{4} Ω = ∮_{\partial V} S \cdot d Σ

for any tensor, such as

S = S^{α}_{β}^{γ} e_{α} \otimes ω^{β} \otimes e_{γ}

(see Box 5.3 for component form). It is an application of the generalized Stokes Theorem (Box 4.1), and depends on the fact that the basis vectors

e_{α}

and

ω^{β}

of a global Lorentz frame are constants, i.e., are independent of

x

. The definitions follow in A ; the proof is in B .
A. Tensor-valued integrals can be defined in flat spaces because one uses constant basis vectors. Thus one defines

\int S \cdot d^{3} Σ = e_{α} \otimes ω^{β} \int S_{β}^{α} d^{3} Σ_{γ}

for a tensor of the indicated rank. One justifies pulling basis vectors and forms outside the integral sign because they are constants, independent of location in spacetime. Each of the numbers

\int S^{α}_{β}^{γ} d^{3} Σ_{γ}

(for

α, β = 0, 1, 2, 3

) is then evaluated by substituting any properly oriented parametrization of the hypersurface into the 3 -form

S^{α}_{β}^{γ} d^{3} Σ_{γ}

as described in Box 4.1 (arbitrary curvilinear parametrization in the part of the calculation not involving the "free indices"

α

and

β

). In other words,

S \cdot d^{3} Σ = e_{α} \otimes ω^{β} \otimes S^{α}_{β}^{γ} d^{3} Σ_{γ}

is considered a "tensor-valued 3-form." Under an integral sign, it is contracted with the hyperplane element tangent to the 3 -surface

P (λ^{1}, λ^{2}, λ^{3})

of integration to form the integral

\begin{aligned} \int S \cdot d^{3} Σ & = \int ⟨ S \cdot d^{3} Σ, \frac{\partial P}{\partial λ^{1}} \land \frac{\partial P}{\partial λ^{2}} \land \frac{\partial P}{\partial λ^{3}} ⟩ d λ^{1} d λ^{2} d λ^{3} \\ = e_{α} \otimes ω^{β} \int S^{α}_{β} γ_{ε_{γ | λ μ ν |}} \underset{\begin{array}{c} Jacobian \\ determinant \end{array}}{\underset{⏟}{\frac{\partial (x^{λ}, x^{μ}, x^{ν})}{\partial (λ^{1}, λ^{2}, λ^{3})}}} d λ^{1} d λ^{2} d λ^{3} . \end{aligned}

Although constant basis vectors

e_{α}, ω^{β}

derived from rectangular coordinates are essential here, a completely general parametrization of the hypersurface may be used.
B. The proof of Gauss's Theorem is a computation:

\begin{aligned} ∮_{\partial γ} s \cdot d^{3} Σ & = e_{α} \otimes ω^{β} ∮_{\partial V} S^{α}_{β}^{γ} d^{3} Σ_{γ} & (e_{α}, ω^{β} are constant) \\ = e_{α} \otimes ω^{β} \int_{γ} d (S^{α}_{β}^{γ} d^{3} Σ_{γ}) & (Stokes Theorem) \\ = e_{α} \otimes ω^{β} \int_{γ} S^{α}_{β}^{γ},^{'}^{*} 1 & (see below) \\ = \int_{γ} (\nabla \cdot S) d^{4} Ω . & (merely notation) \end{aligned}

The missing computational step above is

\begin{aligned} d (S^{α}_{β}^{γ} d^{3} Σ_{γ}) & = (\partial S^{α}_{β}^{γ} / \partial x^{ρ}) d x^{ρ} \land d^{3} Σ_{γ} \\ = {(\partial S^{α}_{β}^{γ} / \partial x^{γ})}^{*} 1 . \end{aligned}

Here the first step uses

d (d^{3} Σ_{γ}) = 0

(which follows from

ε_{μ α β γ} =

const in flat spacetime). The second step uses

d x^{ρ} \land d^{3} Σ_{γ} = δ_{γ}^{ρ} * 1

[Write the lefthand side of this identity as

ε_{γ | μ ν λ |} d x^{ρ} \land d x^{μ} \land d x^{ν} \land d x^{λ}

. The only possible non-zero term in the sum over

μ ν λ

is the one with

μ < ν < λ

all different from

ρ

. The righthand side is the value of this term.]
C. Glossary of notations.

Charge density 3-form:

\begin{aligned} * J = J^{μ} d^{3} Σ_{μ} & = J \cdot d^{3} Σ \\ = \underset{{(^{*} J)}_{α β γ}}{\underset{⏟}{J^{μ} ε_{μ α β γ}}} \underset{d^{3} Σ_{μ}}{\underset{⏟}{}} d x^{α} \land d x^{β} \land d x^{γ} / 3! \end{aligned}

Maxwell and Faraday 2-forms:

\begin{aligned} * F & = \frac{1}{2} F^{μ ν} d^{2} S_{μ ν} \\ F & = \frac{1}{2} F_{μ ν} d x^{μ} \land d x^{ν} . \end{aligned}

Basis 2-forms:

\begin{array}{lr} d x^{α} \land d x^{β}; & (one way to label) \\ d^{2} S_{μ ν} = ε_{μ ν | α β |} d x^{α} \land d x^{β} . & (dual way to label) \end{array}

Energy-momentum density 3-form:

\overset{\equiv e_{μ} T^{μ ν} d^{3} Σ_{ν} \equiv^{*} T;}{\overset{⏞}{dual on last index, {(^{*} T)}^{μ}_{α β γ} = T^{3} T^{μ ν} ε_{ν α β γ}}}

Angular momentum density 3-form:

\begin{aligned} J \cdot d^{3} Σ \equiv \frac{1}{2} e_{μ} \land e_{ν} g^{μ ν α} d^{3} Σ_{α} & \equiv^{*} g \\ {(^{*} g)}^{μ ν}_{α β γ} & = g^{μ ν λ} ε_{λ α β γ} . \end{aligned}

(In the frame-independent equation

\nabla \cdot T = 0

, one need not worry about which slot of

T

to take the divergence on; the slots are symmetric, so either can be used.)

The equation

\nabla \cdot T = 0

is the differential formulation of the law of 4-momentum conservation. It is also called the equation of motion for stress-energy, because it places constraints on the dynamic evolution of the stress-energy tensor. To examine these constraints for simple systems is to realize the beauty and power of the equation

\nabla \cdot T = 0

§5.10. SAMPLE APPLICATIONS OF $\nabla \cdot T = 0$

The equation of motion

\nabla \cdot T = 0

makes contact with the classical (Newtonian) equations of hydrodynamics, when applied to a nearly Newtonian fluid. Such a fluid has low velocities relative to the Lorentz frame used,

| v^{j} | ≪ 1

; and in its rest frame its pressure is small compared to its density of mass-energy,

p / ρ = p / ρ c^{2} ≪ 1

. For example, the air in a hurricane has

\begin{matrix} | v^{j} | \sim 100 km / hour \sim 3, 000 cm / \sec \sim 10^{- 7} c = 10^{- 7} ≪ 1, \\ \frac{p}{ρ} \sim \frac{1 atmosphere}{10^{- 3} g / {cm}^{3}} \sim \frac{10^{6} dynes / {cm}^{2}}{10^{- 3} g / {cm}^{3}} = 10^{9} \frac{{cm}^{2}}{\sec^{2}} \sim 10^{- 12} c^{2} = 10^{- 12} ≪ 1 . \end{matrix}

The stress-energy tensor for such a fluid has components

\begin{aligned} (5.37a) & T^{00} & = (ρ + p) u^{0} u^{0} - p \approx ρ, \\ (5.37b) & T^{0 j} & = T^{j 0} = (ρ + p) u^{0} u^{j} \approx ρ v^{j}, \\ (5.37c) & T^{j k} & = (ρ + p) u^{j} u^{k} + p δ^{j k} \approx ρ v^{j} v^{k} + p δ^{j k}; \end{aligned}

and the equation of motion

\nabla \cdot T = 0

has components

\begin{matrix} (5.38a) & T^{00}_{, 0} + T^{0 j}_{, j} = \partial ρ / \partial t + \nabla \cdot (ρ v) = 0 \end{matrix}

("equation of continuity");
and

T_{, 0}^{j 0} + T^{j k}_{, k} = \partial (ρ v^{j}) / \partial t + \partial (ρ v^{j} v^{k}) / \partial x^{k} + \partial p / \partial x^{j} = 0

or, equivalently (by combining with the equation of continuity),

\begin{matrix} (5.38b) & \frac{\partial v}{\partial t} + (v \cdot \nabla) v = - \frac{1}{ρ} \nabla p ("Euler's equation"). \end{matrix}

Box 5.5 derives and discusses these results from the Newtonian viewpoint.
As a second application of

\nabla \cdot T = 0

, consider a composite system: a block of rubber with electrically charged beads imbedded in it, interacting with an electromagnetic field. The block of rubber vibrates, and its accelerating beads radiate electromagnetic waves; at the same time, incoming electromagnetic waves push on the beads, altering the pattern of vibration of the block of rubber. The interactions shove 4-momentum back and forth between beaded block and electromagnetic field.

Box 5.5 NEWTONIAN HYDRODYNAMICS REVIEWED

Consider a classical, nonrelativistic, perfect fluid. Apply Newton's law

F = m a

to a "fluid particle"; that is, to a small fixed mass of fluid followed in its progress through space:

\frac{d}{d t} (

momentum per unit mass

) = (

force per unit mass

)

= \frac{(force per unit volume)}{(density)} = \frac{- (gradient of pressure)}{(density)}

\begin{matrix} (1) & \frac{d v}{d t} = - \frac{1}{ρ} \nabla p \end{matrix}

Translate from time-rate of change following the fluid to time-rate of change as measured at a fixed location, finding

(\begin{matrix} rate of change \\ with time \\ following fluid \end{matrix}) = (\begin{matrix} rate of change \\ with time at \\ fixed location \end{matrix}) + (\binom{velocity}{of fluid}) \cdot (\binom{rate of change}{with position})

\begin{matrix} (2) & \frac{\partial v}{\partial t} + (v \cdot \nabla) v = - \frac{1}{ρ} \nabla p \end{matrix}

\frac{\partial v_{i}}{\partial t} + v_{i, k} v_{k} = - \frac{1}{ρ} p_{, i} .

(Latin indices run from 1 to 3; summation convention; upper and lower indices used indifferently for space dimensions in flat space!) This is Euler's fundamental equation for the hydrodynamics of a perfect fluid.

Two further equations are needed to complete the description of a perfect fluid. One states the absence of heat transfer by requiring that the specific entropy (entropy per unit mass) be constant for each fluid "particle":

\begin{matrix} (3) & \frac{d s}{d t} = 0, or \frac{\partial s}{\partial t} + (v \cdot \nabla) s = 0 . \end{matrix}

The final equation expresses the conservation of mass:

\begin{matrix} (4) & \frac{\partial ρ}{\partial t} + \nabla \cdot (ρ v) = 0 \end{matrix}

\frac{\partial ρ}{\partial t} + {(ρ v_{k})}_{, k} = 0

Box 5.5 (continued)

it is analagous in every way to the equation that expresses conservation of charge in electrodynamics and that bears the same name, "equation of continuity."

The Newtonian stress-energy tensor, like its relativistic counterpart, is linked to conservation of momentum and mass. Therefore examine the time-rate of change of the density of fluid momentum,

ρ v_{i}

, contained in a unit volume; thus,

\begin{matrix} (5) & \partial (ρ v_{i}) / \partial t = - {(ρ v_{i} v_{k})}_{k} - p_{, i} . \end{matrix}

Momentum flows into the little volume element on the left ("force equals time-rate of change of momentum") and out on the right; similarly at the other faces. Therefore the righthand side of (5) must represent the divergence of this momentum flux:

\begin{matrix} (6) & \partial (ρ v_{i}) / \partial t = - T_{i k, k} \end{matrix}

Consequently, we take for the momentum flux itself

\begin{matrix} (7) & T^{i k} = T_{i k} = \underset{"convection"}{\underset{⏟}{}} \frac{ρ v_{i} v_{k}}{"push"} + \underset{δ_{i k} p}{\underset{⏟}{}} . \end{matrix}

For the momentum density, the Newtonian value is

\begin{matrix} (8) & T^{0 i} = T^{i 0} = ρ v_{i} \end{matrix}

With this notation, the equation for the time-rate of change of momentum becomes

\begin{matrix} (9) & \partial T^{i μ} / \partial x^{μ} = 0; \end{matrix}

and with

T^{00} = ρ

, the equation of continuity reads

\begin{matrix} (10) & \partial T^{0 μ} / \partial x^{μ} = 0 . \end{matrix}

In conclusion, these Newtonian considerations give a reasonable approximation to the relativistic stress-energy tensor:

The 4-momentum of neither block nor field is conserved; neither

\nabla \cdot T_{(block)}

nor

\nabla \cdot T_{(em field)}

vanishes. But total 4-momentum must be conserved, so

\begin{matrix} (5.39) & \nabla \cdot (T_{(block)} + T_{(em field)}) must vanish. \end{matrix}

For a general electromagnetic field interacting with any source,

\nabla \cdot T_{(em field)}

has the form

\begin{matrix} (5.40) & T_{lem field), ν}^{μ ν} = - F^{μ α} J_{α} . \end{matrix}

(This was derived in exercise 3.18 by combining

T^{μ ν}_{, ν} = 0

with expression 5.22 for the electromagnetic stress-energy tensor, and with Maxwell's equations.) For our beaded block,

J

is the 4 -current associated with the vibrating, charged beads, and

F

is the electromagnetic field tensor. The time component of equation (5.40) reads

\begin{aligned} (5.41) & T_{(em field, ν}^{0 ν} & = - F^{0 k} J_{k} = - E \cdot J \\ = - (\binom{rate at which electric field E does work}{on a unit volume of charged beads}) . \end{aligned}

For comparison,

T_{(block), 0}^{00}

is the rate at which the block's energy density changes with time,

- T_{(block), j is}^{0 j}

ithe contribution of the block's energy flux to this rate of change of energy density, and consequently their difference

T_{(block),,}^{0 ν}

, has the meaning

\begin{matrix} (5.42) & T_{(block), v}^{0 v} = (\begin{array}{l} rate at which mass-energy of block per \\ unit volume increases due to actions \\ other than internal mechanical forces \\ between one part of block and another \end{array}) \end{matrix}

Hence, the conservation law

{(T_{(em field)}^{0 ν} + T_{(block)}^{0 ν})}_{, ν} = 0

says that the mass-energy of the block increases at precisely the same rate as the electric field does work on the beads. A similar result holds for momentum:

\begin{matrix} (5.44) & \begin{aligned} T_{(em field), ν}^{k ν} e_{k} & = - F^{k ν} J_{ν} e_{k} = - (J^{0} E + J \times B) \\ = - (\binom{Lorentz force per unit volume}{acting on beads}), \\ T_{(block),,}^{k ν} e_{k} & = (\begin{matrix} rate at which momentum per unit volume \\ of block increases due to actions \\ other than its own stresses \end{matrix} \end{aligned}); \end{matrix}

so the conservation law

{(T_{(em field)}^{k ν} + T_{(block}^{k ν})}_{, p} = 0

says that the rate of change of the momentum of the block equals the force of the electromagnetic field on its beads.

Angular momentum defined and its integral conservation law derived

§5.11. ANGULAR MOMENTUM

The symmetry,

T^{μ ν} = T^{ν μ}

, of the stress-energy tensor enables one to define a conserved angular momentum

J^{α β}

, analogous to the linear momentum

p^{α}

. The angular momentum is defined relative to a specific but arbitrary origin-an event

a

with coordinates, in a particular Lorentz frame,

\begin{matrix} (5.45) & x^{α} (a) = a^{α} . \end{matrix}

The angular momentum about

a

is defined using the tensor

\begin{matrix} (5.46) & g^{α β γ} = (x^{α} - a^{α}) T^{β γ} - (x^{β} - a^{β}) T^{α γ} . \end{matrix}

(Note that

x^{α} - a^{α}

is the vector separation of the "field point"

x^{α}

from the "origin"

A; T^{α γ}

is here evaluated at the "field point".) Because of the symmetry of

T, g^{α β γ}

has vanishing divergence:

\begin{aligned} g^{α β γ}_{, γ} & = δ^{α}_{γ} T^{β γ} + (x^{α} - a^{α}) \underset{0}{\underset{⏟}{T^{β γ}_{γ},}} - δ^{β}_{γ} T^{α γ} - (x^{β} - a^{β}) \underset{0}{\underset{⏟}{T^{α γ}, γ}} \\ (5.47) & = T^{β α} - T^{α β} = 0 . \end{aligned}

Consequently, its integral over any closed 3 -surface vanishes

\begin{matrix} (5.48) & ∮_{\partial γ} g^{α β γ} d^{3} Σ_{γ} = 0 \end{matrix}

("integral form of the law of conservation of angular momentum").
The integral over a spacelike surface of constant time

t

\begin{matrix} (5.49) & J^{α β} = \int g^{α β 0} d x d y d z = \int [(x^{α} - a^{α}) T^{β 0} - (x^{β} - a^{β}) T^{α 0}] d x d y d z \end{matrix}

Recalling that

T^{β 0}

is momentum density, one sees that (5.49) has the same form as the equation "

J = r \times p

" of Newtonian theory. Hence the name "total angular momentum" for

J^{α β}

. Various aspects of this conserved angular momentum, including the tie to its Newtonian cousin, are explored in Box 5.6.

EXERCISES

Exercise 5.2. CHARGE CONSERVATION

Exercise 3.16 revealed that the charge-current 4-vector

J

satisfies the differential conservation law

\nabla \cdot J = 0

. Write down the corresponding integral conservation law, and interpret it for the four closed surfaces of Fig. 5.3.

Exercise 5.3. PARTICLE PRODUCTION

Inside highly evolved, massive stars, the temperature is so high that electron-positron pairs are continually produced and destroyed. Let

S

be the number-flux vector for electrons and positrons, and denote its divergence by

\begin{matrix} (5.50) & ϵ \equiv \nabla \cdot S \end{matrix}

Box 5.6 ANGULAR MOMENTUM

A. Definition of Angular Momentum

(a) Pick an arbitrary spacelike hypersurface

S

and an arbitrary event

a

with coordinates

x^{α} (A) \equiv a^{α}

. (Use globally inertial coordinates throughout.)
(b) Define "total angular momentum on

S

about

a

" to be

\begin{aligned} J^{μ ν} & \equiv \int_{S} g^{μ ν α} d^{3} Σ_{α} \\ g^{μ ν α} & \equiv (x^{μ} - a^{μ}) T^{ν α} - (x^{ν} - a^{ν}) T^{μ α} \end{aligned}

S

is a hypersurface of constant time

t

, this becomes

J^{μ ν} = \int g^{μ ν 0} d x d y d z

B. Conservation of Angular Momentum

(a)

T^{μ ν}_{, ν} = 0

implies

g^{μ ν α}_{, α} = 0

.
(b) This means that

J^{μ ν}

is independent of the hypersurface

S

on which it is calculated (Gauss's theorem):

\begin{aligned} J^{μ ν} (S_{A}) - J^{μ ν} (S_{B}) \\ = \int_{\partial V} g^{μ ν α} d^{3} Σ_{α} \\ = \int_{V} G^{μ ν α}_{, α} d^{4} x = 0 . \end{aligned}

(Note:

\partial V \equiv

(boundary of

V

) includes

S_{A}, S_{B}

, and timelike surfaces at spatial infinity; contribution of latter dropped-localized source.)

C. Change of Point About Which Angular Momentum is Calculated

Let

b^{α}

be vector from

a_{0}

a_{1} : b^{α} = a_{1}^{α} - a_{0}^{α}

. Then

\begin{aligned} J^{μ ν} (about a_{1}) - J^{μ ν} (about a_{0}) \\ = - b^{μ} \int_{S} T^{ν α} d^{3} Σ_{α} + b^{ν} \int_{S} T^{μ α} d^{3} Σ_{α} \\ = - b^{μ} P^{ν} + b^{ν} P^{μ} \end{aligned}

where

P^{μ}

is total 4-momentum.

Box 5.6 (continued)
D. Intrinsic Angular Momentum
(a) Work, for a moment, in the system's rest frame, where

P^{0} = M, P^{j} = 0, x_{C M}^{j} = \frac{1}{M} \int x^{j} T^{00} d^{3} x =

location of center of mass.

Intrinsic angular momentum is defined as angular momentum about any event (

a^{0}, x_{C M}^{j}

) on center of mass's world line. Its components are denoted

S^{μ ν}

and work out to be

S^{0 j} = 0, S^{j k} = ϵ^{j k ℓ} S^{ℓ},

where

\begin{aligned} S & \equiv \int (x - x_{C M}) \times (momentum density) d^{3} x \\ \equiv "intrinsic angular momentum vector." \end{aligned}

(b) Define "intrinsic angular momentum 4 -vector"

S^{μ}

to be that 4 -vector whose components in the rest frame are

(0, S)

; then the above equations say

\begin{aligned} S^{μ ν} & = U_{α} S_{β} ϵ^{α β μ ν}, \\ U_{β} & \equiv P_{β} / M = 4 -velocity of center of mass, \\ U_{β} S^{β} & = 0 . \end{aligned}

E. Decomposition of Angular Momentum into Intrinsic and Orbital Parts

(a) Pick an arbitrary event

a

, whose perpendicular displacement from center-of-mass world line is

- Y^{α}

, so

U_{β} Y^{β} = 0 .

(b) Then, by Part C, the angular momentum about

a

J^{μ ν} = \underset{S^{μ ν} (intrinsic)}{\underset{⏟}{U_{α} S_{β} ϵ^{α β μ ν}}} + \underset{L^{μ ν} (orbital)}{\underset{⏟}{Y^{μ} P^{ν} - Y^{ν} P^{μ}}}

a

, and the 4 -momentum (and hence 4 -velocity), one can calculate the vector from

a

to the center-of-mass world line,

Y^{μ} = - J^{μ ν} P_{ν} / M^{2},

and the intrinsic angular momentum

S_{ρ} = \frac{1}{2} U^{σ} J^{μ ν} ϵ_{σ μ ν ρ} .

Use Gauss's theorem to show that

ϵ

is the number of particles created (minus the number destroyed) in a unit four-dimensional volume of spacetime.

Exercise 5.4. INERTIAL MASS PER UNIT VOLUME

Consider a stressed medium in motion with ordinary velocity

| v | ≪ 1

with respect to a specific Lorentz frame.
(a) Show by Lorentz transformations that the spatial components of the momentum density are

\begin{matrix} (5.51) & T^{0 j} = \sum_{k} m^{j k} v^{k} \end{matrix}

where

\begin{matrix} (5.52) & m^{j k} = T^{\overset{―}{0} \overset{―}{0}} δ^{j k} + T^{\overset{―}{j k}} \end{matrix}

and

T^{\bar{μ} \bar{ν}}

are the components of the stress-energy tensor in the rest frame of the medium. Throughout the solar system

T^{\overset{―}{00}} ≫ | T^{3 k} |

(see, e.g., discussion of hurricane in

§ 5.10

), so one is accustomed to write

T^{0 j} = T^{\overset{―}{00}} v^{j}

, i.e., "

(

momentum density

) = (

rest-mass density

) \times (

velocity)". But inside a neutron star

T^{\overset{―}{0} \overset{―}{0}}

may be of the same order of magnitude as

T^{j \bar{k}}

, so one must replace "(momentum density)

=

(rest-mass density)

\times

(velocity)" by equations (5.51) and (5.52), at low velocities.
(b) Derive equations ( 5.51 ) and (5.52) from Newtonian considerations plus the equivalence of mass and energy. (Hint: the total mass-energy carried past the observer by a volume

V

of the medium includes both the rest mass

T^{\overset{―}{00}} V

and the work done by forces acting across the volume's faces as they "push" the volume through a distance.)
(c) As a result of relation (5.51), the force per unit volume required to produce an acceleration

d v^{k} / d t

in a stressed medium, which is at rest with respect to the man who applies the force, is

\begin{matrix} (5.53) & F^{j} = d T^{0 j} / d t = \sum_{k} m^{j k} d v^{k} / d t \end{matrix}

This equation suggests that one call

m^{j k}

the "inertial mass per unit volume" of a stressed medium at rest. In general

m^{i k}

is a symmetric 3-tensor. What does it become for the special case of a perfect fluid?
(d) Consider an isolated, stressed body at rest and in equilibrium (

T^{α β}_{, 0} = 0

) in the laboratory frame. Show that its total inertial mass, defined by

\begin{matrix} (5.54) & M^{i j} = \int_{\begin{matrix} stressed \\ body \end{matrix}} m^{i j} d x d y d z \end{matrix}

is isotropic and equals the rest mass of the body

\begin{matrix} (5.55) & M^{i j} = δ^{i j} \int T^{n 0} d x d y d z \end{matrix}

Exercise 5.5. DETERMINANTS AND JACOBIANS

(a) Write out explicitly the sum defining

d^{2} S_{01}

d^{2} S_{μ ν} \equiv ϵ_{μ ν α β} \frac{\partial x^{α}}{\partial a} \frac{\partial x^{β}}{\partial b} d a d b .

Thereby establish the formula

d^{2} S_{μ ν} = ϵ_{μ ν | α β |} \frac{\partial (x^{α}, x^{β})}{\partial (a, b)} d a d b = \frac{1}{2!} ϵ_{μ ν α β} \frac{\partial (x^{α}, x^{β})}{\partial (a, b)} d a d b .

(Expressions such as these should occur only under integral signs. In this exercise one may either supply an

\int \dots

wherever necessary, or else interpret the differentials in terms of the exterior calculus,

d a d b ⟶ d a \land d b

; see Box 5.4.) The notation used here for Jacobian determinants is

\frac{\partial (f, g)}{\partial (a, b)} = | \begin{array}{ll} \frac{\partial f}{\partial a} & \frac{\partial f}{\partial b} \\ \frac{\partial g}{\partial a} & \frac{\partial g}{\partial b} \end{array} |

(b) By a similar inspection of a specific case, show that

d^{3} Σ_{μ} \equiv ϵ_{μ α β γ} \frac{\partial x^{α}}{\partial a} \frac{\partial x^{β}}{\partial b} \frac{\partial x^{γ}}{\partial c} d a d b d c = \frac{1}{3!} ϵ_{μ α β γ} \frac{\partial (x^{α}, x^{β}, x^{γ})}{\partial (a, b, c)} d a d b d c

(c) Cite a precise definition of the value of a determinant as a sum of terms (with suitably alternating signs), with each term a product containing one factor from each row and simultaneously one factor from each column. Show that this definition can be stated (in the

4 \times 4

case, with the

p \times p

case an obvious extension) as

\det A \equiv \det ‖ A^{λ}_{ρ} ‖ = ϵ_{α β γ δ} A^{α}_{0} A^{β}_{1} A^{γ}_{2} A^{8}_{3} .

(d) Show that

\det A = \frac{1}{4!} δ_{α β γ δ}^{μ ν ρ σ} A^{α}_{μ} A^{β}_{ν} A^{γ}_{ρ} A^{δ}_{σ}

(for a definition of

δ_{α β γ δ}^{μ ν ρ σ}

, see exercises 3.13 and 4.12).
(e) Use properties of the

δ

-symbol to show that the matrix

A^{- 1}

inverse to

A

has entries

{(A^{- 1})}^{μ}_{α}

given by

{(A^{- 1})}^{μ}_{α} (\det A) = \frac{1}{3!} δ_{α β γ δ}^{μ ν ρ σ} A^{β}_{ν} A^{γ}_{ρ} A^{δ}_{σ} .

(f) By an "index-mechanics" computation, from the formula for

\det A

in part (d) derive the following expression for the derivative of the logarithm of the determinant

d \ln | \det A | = trace (A^{- 1} d A)

Here

d A

is the matrix

‖ d A^{α}_{μ} ‖

whose entries are 1-forms.

Exercise 5.6. CENTROIDS AND SIZES

Consider an isolated system with stress-energy tensor

T^{μ ν}

, total 4-momentum

P^{α}

, magnitude of 4 -momentum

M = (- P \cdot P)^{1 / 2}

, intrinsic angular momentum tensor

S^{α β}

, and intrinsic angular momentum vector

S^{α}

. (See Box 5.6.) An observer with 4 -velocity

u^{α}

defines the centroid of the system, at his Lorentz time

x^{0} = t

and in his own Lorentz frame, by

\begin{matrix} (5.56) & X_{u}^{j} (t) = (1 / P^{0}) \int_{x^{0} = t} x^{j} T^{00} d^{3} x in Lorentz frame where u = \partial P / \partial x^{0} \end{matrix}

This centroid depends on (i) the particular system being studied, (ii) the 4 -velocity

u

of the observer, and (iii) the time

t

at which the system is observed.
(a) Show that the centroid moves with a uniform velocity

\begin{matrix} (5.57) & d X_{u}^{j} / d t = P^{j} / P^{0} \end{matrix}

corresponding to the 4 -velocity

\begin{matrix} (') & U = P / M \end{matrix}

Note that this " 4 -velocity of centroid" is independent of the 4 -velocity

u

used in defining the centroid.
(b) The centroid associated with the rest frame of the system (i.e., the centroid defined with

u = U

) is called the center of mass; see Box 5.6 . Let

ξ_{u}

be a vector reaching from any event on the center-of-mass world line to any event on the world line of the centroid associated with 4 -velocity

u

; thus the components of

ξ_{u}

in any coordinate system are

\begin{matrix} (5.58) & ξ_{u}^{α} = X_{u}^{α} - X_{U}^{α} \end{matrix}

Show that

ξ_{u}

satisfies the equation

\begin{matrix} (5.59) & [(ξ_{u}^{α} P^{β} - P^{α} ξ_{u}^{β}) - S^{α β}] u_{β} = 0 \end{matrix}

[Hint: perform the calculation in a Lorentz frame where

u = \partial P / \partial x^{0}

.]
(c) Show that, as seen in the rest-frame of the system at any given moment of time, the above equation reduces to the three-dimensional Euclidean equation

\begin{matrix} (') & ξ_{u} = - (v \times S) / M \end{matrix}

where

v = u / u^{0}

is the ordinary velocity of the frame associated with the centroid.
(d) Assume that the energy density measured by any observer anywhere in spacetime is
non-negative

(u \cdot T \cdot u \geq 0

for all timelike

u)

. In the rest frame of the system, construct the smallest possible cylinder that is parallel to

S

and that contains the entire system (

T_{α β} = 0

everywhere outside the cylinder). Show that the radius

r_{0}

of this cylinder is limited by

\begin{matrix} (5.60) & r_{0} \geq | S | / M \end{matrix}

Thus, a system with given intrinsic angular momentum

S

and given mass

M

has a minimum possible size

r_{0 min} = | S | / M

as measured in its rest frame.

синотет 6

ACCELERATED OBSERVERS

The objective world simply is; it does not happen. Only to the gaze of my consciousness, crawling upward along the life line [world line] of my body, does a section of this world come to life as a fleeting image in space which continuously changes in time.

§6.1. ACCELERATED OBSERVERS CAN BE ANALYZED USING SPECIAL RELATIVITY

It helps in analyzing gravitation to consider a situation where gravity is mocked up by acceleration. Focus attention on a region so far from any attracting matter, and so free of disturbance, that (to some proposed degree of precision) spacetime there can be considered to be flat and to have Lorentz geometry. Let the observer acquire the feeling that he is subject to gravity, either because of jet rockets strapped to his legs or because he is in a rocket-driven spaceship. How does physics look to him?

Dare one answer this question? At this early stage in the book, is one not too ignorant of gravitation physics to predict what physical effects will be measured by an observer who thinks he is in a gravitational field, although he is really in an accelerated spaceship? Quite the contrary; special relativity was developed precisely to predict the physics of accelerated objects-e.g., the radiation from an accelerated charge. Even the fantastic accelerations

a_{nuclear} \sim v^{2} / r \sim 10^{31} cm / \sec^{2} \sim 10^{28} "earth gravities"

suffered by a neutron bound in a nucleus, and the even greater accelerations met in high-energy particle-scattering events, are routinely and accurately treated within

Accelerated motion and accelerated observers can be analyzed using special relativity

Box 6.1 GENERAL RELATIVITY IS BUILT ON SPECIAL RELATIVITY

A tourist in a powered interplanetary rocket feels "gravity." Can a physicist by local effects convince him that this "gravity" is bogus? Never, says Einstein's principle of the local equivalence of gravity and accelerations. But then the physicist will make no errors if he deludes himself into treating true gravity as a local illusion caused by acceleration. Under this delusion, he barges ahead and solves gravitational problems by using special relativity: if he is clever enough to divide every problem into a network of local questions, each solvable under such a delusion, then he can work out all influ-
ences of any gravitational field. Only three basic principles are invoked: special-relativity physics, the equivalence principle, and the local nature of physics. They are simple and clear. To apply them, however, imposes a double task: (1) take spacetime apart into locally flat pieces (where the principles are valid), and (2) put these pieces together again into a comprehensible picture. To undertake this dissection and reconstitution, to see curved dynamic spacetime inescapably take form, and to see the consequences for physics: that is general relativity.
the framework of special relativity. The theoretician who confidently applies special relativity to antiproton annihilations and strange-particle resonances is not about to be frightened off by the mere illusions of a rocket passenger who gullibly believed the travel brochures advertising "earth gravity all the way." When spacetime is flat, move however one will, special relativity can handle the job. (It can handle bigger jobs too; see Box 6.1.) The essential features of how special relativity handles the job are summarized in Box 6.2 for the benefit of the Track-1 reader, who can skip the rest of the chapter, and also for the benefit of the Track- 2 reader, who will find it useful background for the rest of the chapter.

Box 6.2 ACCELERATED OBSERVERS IN BRIEF

An accelerated observer can carry clocks and measuring rods with him, and can use them to set up a reference frame (coordinate system) in his neighborhood.
His clocks, if carefully chosen so their structures are affected negligibly by acceleration (e.g., atomic clocks), will tick at the same rate as unaccelerated clocks moving momentarily along with him:

Δ τ \equiv (\begin{matrix} time interval ticked off \\ by observer's clocks as he \\ moves a vector displacement \\ ξ along his world line \end{matrix}) = [- g (ξ, ξ)]^{1 / 2}

And his rods, if chosen to be sufficiently rigid, will measure the same lengths as
momentarily comoving, unaccelerated rods do. (For further discussion, see

§ 16.4

, and Boxes 16.2 to 16.4.)
Let the observer's coordinate system be a Cartesian latticework of rods and clocks, with the origin of the lattice always on his world line. He must keep his latticework small:

l \equiv (\binom{spatial dimensions}{of lattice}) ≪ {(\binom{the acceleration measured}{by accelerometers he carries})}^{- 1} \equiv \frac{1}{g} .

At distances

l

away from his world line, strange things of dimensionless magnitude

g l

happen to his lattice-e.g., the acceleration measured by accelerometers differs from

g

by a fractional amount

\sim g l

(exercise 6.7); also, clocks initially synchronized with the clock on his world line get out of step (tick at different rates) by a fractional amount

\sim g l

(exercise 6.6). (Note that an acceleration of one "earth gravity" corresponds to

g^{- 1} \sim 10^{- 3} \sec^{2} / cm \sim 10^{18} cm \sim 1 light-year

so the restriction

l ≪ 1 / g

is normally not severe.)
To deduce the results of experiments and observations performed by an accelerated observer, one can analyze them in coordinate-independent, geometric terms, and then project the results onto the basis vectors of his accelerated frame. Alternatively, one can analyze the experiments and observations in a Lorentz frame, and then transform to the accelerated frame.
As deduced in this manner, the results of experiments performed locally (at

l ≪ 1 / g)

by an accelerated observer differ from the results of the same experiments performed in a Lorentz frame in only three ways:
(1) There are complicated fractional differences of order

g l ≪ 1

mentioned above, that can be made negligible by making the accelerated frame small enough.
(2) There are Coriolis forces of precisely the same type as are encountered in Newtonian theory (exercise 6.8). These the observer can get rid of by carefully preventing his latticework from rotating-e.g., by tying it to gyroscopes that he accelerates with himself by means of forces applied to their centers of mass (no torque!). Such a nonrotating latticework has "Fermi-Walker transported" basis vectors (§6.5),

\begin{matrix} (1) & \frac{d e_{α^{'}}}{d τ} = u (a \cdot e_{α^{'}}) - a (u \cdot e_{α^{'}}), \end{matrix}

where

u = 4

-velocity, and

a = d u / d τ = 4

-acceleration.
(3) There are inertial forces of precisely the same type as are encountered in Newtonian theory (exercise 6.8). These are due to the observer's acceleration, and he cannot get rid of them except by stopping his accelerating.

§6.2. HYPERBOLIC MOTION

Study a rocket passenger who feels "gravity" because he is being accelerated in flat spacetime. Begin by describing his motion relative to an inertial reference frame. His 4 -velocity satisfies the condition

u^{2} = - 1

. To say that it is fixed in magnitude is to say that the 4 -acceleration,

\begin{matrix} (6.1) & a = d u / d τ \end{matrix}

is orthogonal to the 4 -velocity:

\begin{matrix} (6.2) & 0 = (d / d τ) (- 1 / 2) = (d / d τ) (\frac{1}{2} u \cdot u) = a \cdot u \end{matrix}

This equation implies that

a^{0} = 0

in the rest frame of the passenger (that Lorentz frame, where, at the instant in question,

u = e_{0}

); in this frame the space components of

a^{μ}

reduce to the ordinary definition of acceleration,

a^{i} = d^{2} x^{i} / d t^{2}

. From the components

a^{μ} = (0; a^{i})

in the rest frame, then, one sees that the magnitude of the acceleration in the rest frame can be computed as the simple invariant

a^{2} = a^{μ} a_{μ} = {(d^{2} x / d t^{2})}^{2} as measured in rest frame

Consider, for simplicity, an observer who feels always a constant acceleration

g

. Take the acceleration to be in the

x^{1}

direction of some inertial frame, and take

x^{2} = x^{3} = 0

. The equations for the motion of the observer in that inertial frame become

\begin{matrix} (6.3) & \frac{d t}{d τ} = u^{0}, \frac{d x}{d τ} = u^{1}; \frac{d u^{0}}{d τ} = a^{0}, \frac{d u^{1}}{d τ} = a^{1} \end{matrix}

Write out the three algebraic equations

\begin{aligned} u^{μ} u_{μ} = - 1 \\ u^{μ} a_{μ} = - u^{0} a^{0} + u^{1} a^{1} = 0 \end{aligned}

and

a^{μ} a_{μ} = g^{2} .

Solve for the acceleration, finding

\begin{matrix} (6.4) & a^{0} = \frac{d u^{0}}{d τ} = g u^{1}, a^{1} = \frac{d u^{1}}{d τ} = g u^{0} . \end{matrix}

These linear differential equations can be solved immediately. The solution, with a suitable choice of the origin, reads

\begin{matrix} (6.5) & t = g^{- 1} \sinh g τ, x = g^{- 1} \cosh g τ \end{matrix}

Note that

x^{2} - t^{2} = g^{- 2}

. The world line is a hyperbola in a spacetime diagram ("hyperbolic motion"; Figure 6.1). Several interesting aspects of this motion are

Figure 6.1.
Hyperbolic motion. World line of an object that (or an observer who) experiences always a fixed acceleration

g

with respect to an inertial frame that is instantaneously comoving (different inertial frames at different instants!). The 4 -acceleration

a

is everywhere orthogonal (Lorentz geometry!) to the 4-velocity

u

.
treated in the exercises. Let the magnitude of the constant acceleration

g

be the acceleration of gravity,

g = 980 cm / \sec^{2}

experienced on earth:

g ≃ (10^{3} cm / \sec^{2})

{(3 \times 10^{10} cm / \sec)}^{2} = {(3 \times 10^{7} \sec \cdot 3 \times 10^{10} cm / \sec)}^{- 1} = (1 light-year)^{- 1}

. Thus the observer will attain relativistic velocities after maintaining this acceleration for something like one year of his own proper time. He can outrun a photon if he has a head start on it of one light-year or more.

Exercise 6.1. A TRIP TO THE GALACTIC NUCLEUS

Compute the proper time required for the occupants of a rocket ship to travel the

\sim 30, 000

light-years from the Earth to the center of the Galaxy. Assume that they maintain an acceleration of one "earth gravity"

(10^{3} cm / \sec^{2})

for half the trip, and then decelerate at one earth gravity for the remaining half.

Exercise 6.2. ROCKET PAYLOAD

What fraction of the initial mass of the rocket can be payload for the journey considered in exercise 6.1? Assume an ideal rocket that converts rest mass into radiation and ejects all the radiation out the back of the rocket with 100 per cent efficiency and perfect collimation.

Exercise 6.3. TWIN PARADOX

(a) Show that, of all timelike world lines connecting two events

C

and

B

, the one with the longest lapse of proper time is the unaccelerated one. (Hint: perform the calculation in the inertial frame of the unaccelerated world line.)
(b) One twin chooses to move from

C

B

along the unaccelerated world line. Show that the other twin, by an appropriate choice of accelerations, can get from

C

B

in arbitrarily small proper time.
(c) If the second twin prefers to ride in comfort, with the acceleration he feels never exceeding one earth gravity,

g

, what is the shortest proper time-lapse he can achieve between

a

and

B

? Express the answer in terms of

g

and the proper time-lapse

Δ τ

measured by the unaccelerated twin.
(d) Evaluate the answer numerically for several interesting trips.

Exercise 6.4. RADAR SPEED INDICATOR

A radar set measures velocity by emitting a signal at a standard frequency and comparing it with the frequency of the signal reflected back by another object. This redshift measurement is then converted, using the standard special-relativistic formula, into the corresponding velocity, and the radar reads out this velocity. How useful is this radar set as a velocity-measuring instrument for a uniformly accelerated observer?
(a) Consider this problem first for the special case where the object and the radar set are at rest with respect to each other at the instant the radar pulse is reflected. Compute the redshift

1 + z = ω_{e} / ω_{0}

that the radar set measures in this case, and the resulting (incorrect) velocity it infers. Simplify by making use of the symmetries of the situation.
(b) Now consider the situation where the object has a non-zero velocity in the momentary rest frame of the observer at the instant it reflects the radar pulse. Compute the ratio of the actual relative velocity to the velocity read out by the radar set.

Exercise 6.5. RADAR DISTANCE INDICATOR

Use radar as a distance-measuring device. The radar set measures its proper time

τ

between the instant at which it emits a pulse and the later instant when it receives the reflected pulse. It then performs the simple computation

L_{0} = τ / 2

and supplies as output the "distance"

L_{0}

. How accurate is the output reading of the radar set for measuring the actual distance

L

to the object, when used by a uniformly accelerated observer? (

L

is defined as the distance in the momentary rest frame of the observer at the instant the pulse is reflected, which is at the observer's proper time halfway between emitting and receiving the pulse.) Give a correct formula relating

L_{0} \equiv τ / 2

to the actual distance

L

. Show that the reading

L_{0}

becomes infinite as

L

approaches

g^{- 1}

, where

g

is the observer's acceleration, as measured by an accelerometer he carries.

Difficulties in constructing "the coordinate system of an accelerated observer":

Breakdown in communication between observer and events at distance

l >

(acceleration

^{- 1}

§6.3. CONSTRAINTS ON SIZE OF AN ACCELERATED FRAME

It is very easy to put together the words "the coordinate system of an accelerated observer," but it is much harder to find a concept these words might refer to. The most useful first remark one can make about these words is that, if taken seriously, they are self-contradictory. The definite article "the" in this phrase suggests that one is thinking of some unique coordinate system naturally associated with some specified accelerated observer, such as one whose world line is given in equation (6.5). If the coordinate system is indeed natural, one would expect that the coordinates of any event could be determined by a sufficiently ingenious observer by sending and receiving light signals. But from Figure 6.1 it is clear that the events composing one quarter of all spacetime (Zone III) can neither send light signals to, nor receive light signals from, the specified observer. Another half of spacetime suffers lesser disabilities in this respect: Zone II cannot send to the observer, Zone IV cannot receive from him. It is hard to see how the observer could define in any natural way a coordinate system covering events with which he has no causal relationship, which he cannot see, and from which he cannot be seen!

Difficulties also occur when one considers an observer who begins at rest in one frame, is accelerated for a time, and maintains thereafter a constant velocity, at rest in some other inertial coordinate system. Do his motions define in any natural way

Figure 6.2.
World line of an observer who has undergone a brief period of acceleration. In each phase of motion at constant velocity, an inertial coordinate system can be set up. However, there is no way to reconcile these discordant coordinates in the region of overlap (beginning at distance

g^{- 1}

to the left of the region of acceleration).
a coordinate system? Then this coordinate system (1) should be the inertial frame

x^{μ}

in which he was at rest for times

x^{0}

less than 0 , and (2) should be the other inertial frame

x^{μ^{'}}

for times

x^{0^{'}} > T^{'}

in which he was at rest in that other frame. Evidently some further thinking would be required to decide how to define the coordinates in the regions not determined by these two conditions (Figure 6.2). More serious, however, is the fact that these two conditions are inconsistent for a region of spacetime that satisfies simultaneously

x^{0} < 0

and

x^{0^{'}} > T^{'}

. In both examples of accelerated motion (Figures 6.1 and 6.2), the serious difficulties about defining a coordinate system begin only at a finite distance

g^{- 1}

from the world line of the accelerated observer. The problem evidently has no solution for distances from the world line greater than

g^{- 1}

. It does possess a natural solution in the immediate vicinity of the observer. This solution goes under the name of "Fermi-Walker transported orthonormal tetrad." The essential idea lends itself to simple illustration for hyperbolic motion, as follows.

§6.4. THE TETRAD CARRIED BY A UNIFORMLY ACCELERATED OBSERVER

An infinitesimal version of a coordinate system is supplied by a "tetrad," or "moving frame" (Cartan's "repère mobile"), or set of basis vectors

e_{0^{'}}, e_{1^{'}}, e_{2^{'}}, e_{3^{'}}

(subscript tells which vector, not which component of one vector!) Let the time axis be the time axis of a comoving inertial frame in which the observer is momentarily at rest. Thus the zeroth basis vector is identical with his 4-velocity:

e_{0^{'}} = u

. The space axes

e_{2}

and

e_{3}

are not affected by Lorentz transformations in the 1-direction. Therefore take

e_{2^{'}}

and

e_{3^{'}}

to be the unit basis vectors of the all-encompassing Lorentz frame relative to which the hyperbolic motion of the observer has already been described in equations (6.5):

e_{2^{'}} = e_{2}; e_{3^{'}} = e_{3}

. The remaining basis vector,

e_{1^{'}}

, orthogonal to the other three, is parallel to the acceleration vector,

e_{1^{'}} = g^{- 1} a

[see equation (6.4)]. There is a more satisfactory way to characterize this moving frame: the time axis

e_{0^{'}}

is the observer's 4 -velocity, so he is always at rest in this frame; and the

Natural coordinates inconsistent at distance

l > (acceleration)^{- 1}

Orthonormal tetrad of basis vectors carried by uniformly accelerated observer
other three vectors

e_{1^{'}}

are chosen in such a way as to be (1) orthogonal and (2) nonrotating. These basis vectors are:

\begin{aligned} {(e_{0^{'}})}^{μ} = (\cosh g τ; \sinh g τ, 0, 0); \\ {(e_{1^{'}})}^{μ} = (\sinh g τ; \cosh g τ, 0, 0); \\ {(e_{2^{2}})}^{μ} = (0; 0, 1, 0); \\ (6.6) & {(e_{3^{3}})}^{μ} = (0; 0, 0, 1) . \end{aligned}

There is a simple prescription to obtain these four basis vectors. Take the four basis vectors

e_{0}, e_{1}, e_{2}, e_{3}

of the original global Lorentz reference frame, and apply to them a simple boost in the 1-direction, of such a magnitude that

e_{0^{'}}

comes into coincidence with the 4 -velocity of the observer. The fact that these vectors are all orthogonal to each other and of unit magnitude is formally stated by the equation

\begin{matrix} (6.7) & e_{μ^{'}} \cdot e_{ν^{'}} = η_{μ^{'} ν^{'}} \end{matrix}

§6.5. THE TETRAD FERMI-WALKER TRANSPORTED BY AN OBSERVER WITH ARBITRARY ACCELERATION

Turn now from an observer, or an object, in hyperbolic motion to one whose acceleration, always finite, varies arbitrarily with time. Here also we impose three criteria on the moving, infinitesimal reference frame, or tetrad: (1) the basis vectors

e_{μ^{'}}

of the tetrad must remain orthonormal [equation (6.7)]; (2) the basis vectors must form a rest frame for the observer at each instant

(e_{0^{'}} = u)

; and (3) the tetrad should be "nonrotating."
This last criterion requires discussion. The basis vectors of the tetrad at any proper time

τ

must be related to the basis vectors

e_{0}, e_{1}, e_{2}, e_{3}

of some given inertial frame by a Lorentz transformation

e_{μ^{'}} (τ) = Λ^{ν}_{μ^{'}} (τ) e_{p^{'}}

. Therefore the basis vectors at two successive instants must also be related to each other by a Lorentz transformation. But a Lorentz transformation can be thought of as a "rotation" in spacetime. The 4 -velocity

u

, always of unit magnitude, changes in direction. The very concept of acceleration therefore implies "rotation" of velocity 4 -vector. How then is the requirement of "no rotation" to be interpreted? Demand that the tetrad

e_{μ^{'}} (τ)

change from instant to instant by precisely that amount implied by the rate of change of

u = e_{0}

, and by no additional arbitrary rotation. In other words, (1) accept the inevitable pseudorotation in the timelike plane defined by the velocity 4 -vector and the acceleration, but (2) rule out any ordinary rotation of the three space vectors.
Nonrelativistic physics describes the rotation of a vector (components

v_{i}

) by an instantaneous angular velocity vector (components

ω_{i}

). This angular velocity appears in the formula for the rate of change of

v

\begin{matrix} (6.8) & (d v_{i} / d t) = (ω \times v)_{i} = ε_{i j k} ω_{j} v_{k} \end{matrix}

For the extension to four-dimensional spacetime, it is helpful to think of the rotation
as occurring in the plane perpendicular to the angular velocity vector

ω

. Thus rewrite (6.8) as

\begin{matrix} (6.9) & d v_{i} / d t = - Ω_{i k} v_{k} \end{matrix}

where

\begin{matrix} (6.10) & Ω_{j k} = - Ω_{k j} = ω_{i} ε_{i j k} \end{matrix}

has non-zero components only in the plane of the rotation. In other words, to speak of "a rotation in the

(1, 2)

-plane" is more useful than to speak of a rotation about the 3 -axis. The concept of "plane of rotation" carries over to four dimensions. There a rotation in the

(1, 2)

-plane will leave constant not only the

v_{3}

but also the

v_{0}

component of the velocity. The four-dimensional definition of a rotation is

\begin{matrix} (6.11) & \frac{d v^{μ}}{d τ} = - Ω^{μ ν} v_{ν}, with Ω^{μ ν} = - Ω^{ν μ} . \end{matrix}

To test the appropriateness of this definition of a generalized rotation or infinitesimal Lorentz transformation, verify that it leaves invariant the length of the 4 -vector:

\begin{matrix} (6.12) & d (v_{μ} u^{μ}) / d τ = 2 v_{μ} (d v^{μ} / d τ) = - 2 Ω^{μ ν} v_{μ} v_{ν} = 0 \end{matrix}

The last expression vanishes because

Ω^{μ ν}

is antisymmetric, whereas

v_{μ} v_{ν}

is symmetric. Note also that the antisymmetric tensor

Ω^{μ ν}

("rotation matrix"; "infinitesimal Lorentz transformation") has

4 \times 3 / 2 = 6

independent components. This number agrees with the number of components in a finite Lorentz transformation (three parameters for rotations, plus three parameters for the components of a boost). The "infinitesimal Lorentz transformation" here must (1) generate the appropriate Lorentz transformation in the timelike plane spanned by the 4 -velocity and the 4 -acceleration, and (2) exclude a rotation in any other plane, in particular, in any spacelike plane. The unique answer to these requirements is

\begin{matrix} (6.13) & Ω^{μ ν} = a^{μ} u^{ν} - a^{ν} u^{μ}; i.e., Ω = a \land u . \end{matrix}

Apply this rotation to a spacelike vector

w

orthogonal to

u

and

a, (u \cdot w = 0

and

a \cdot w = 0

). Immediately compute

Ω^{μ ν} w_{v} = 0

. Thus verify the absence of any space rotation. Now check the over-all normalization of

Ω^{μ ν}

in equation (6.13). Apply the infinitesimal Lorentz transformation to the velocity 4 -vector

u

of the observer. Thus insert

v^{μ} = u^{μ}

in (6.11). It then reads

d u^{μ} / d τ \equiv a^{μ} = u^{μ} (a^{ν} u_{ν}) - a^{μ} (u^{ν} u_{ν}) = a^{μ} .

This result is an identity, since

u \cdot u = - 1

and

u \cdot a = 0

.
A vector

v

that undergoes the indicated infinitesimal Lorentz transformation,

\begin{matrix} (6.14) & d v^{μ} / d τ = (u^{μ} a^{ν} - u^{ν} a^{μ}) v_{p} \end{matrix}

Fermi-Walker law of transport for "nonrotating" tetrad of basis vectors carried by an accelerated observer
is said to experience "Fermi-Walker transport" along the world line of the observer.

Mathematics of rotation in spacetime

Figure 6.3.
Construction of spacelike hyperplanes (dashed) orthogonal to the world line (heavy line) of an accelerated particle at selected moments along that world line. Note crossing of hyperplanes at distance

g^{- 1} (τ)

(time-dependent acceleration!) from the world line.

The natural moving frame associated with an accelerated observer consists of four orthonormal vectors, each of which is Fermi-Walker transported along the world line and one of which is

e_{0^{'}} = u

(the 4 -velocity of the observer). Fermi-Walker transport of the space basis vectors

e_{j^{'}}

can be achieved in practice by attaching them to gyroscopes (see Box 6.2 and exercise 6.9).

§6.6. THE LOCAL COORDINATE SYSTEM OF AN ACCELERATED OBSERVER

Tetrad used to construct "local coordinate system of accelerated observer"
"coordinates relative to the accelerated observer." In detail, the prescription for the determination of these four coordinates consists of the four equations

\begin{matrix} (6.16) & x^{μ} = ξ^{k^{'}} {[e_{k^{'}} (τ)]}^{μ} + z^{μ} (τ), \end{matrix}

in which the

x^{μ}

are considered as known, and the coordinates

τ, ξ^{k^{'}}

are considered unknowns.

At a certain distance from the accelerated world line, successive spacelike hyperplanes, instead of advancing with increasing

τ

, will be retrogressing. At this distance, and at greater distances, the concept of "coordinates relative to the accelerated observer" becomes ambiguous and has to be abandoned. To evaluate this distance, note that any sufficiently short section of the world line can be approximated by a hyperbola ("hyperbolic motion with acceleration

g

"), where the time-dependent acceleration

g (τ)

is given by the equation

g^{2} = a^{μ} a_{μ}

Apply the above general prescription to hyperbolic motion, arriving at the equations

\begin{aligned} x^{0} = (g^{- 1} + ξ^{1}) \sinh (g ξ^{0^{'}}), \\ x^{1} = (g^{- 1} + ξ^{1^{1}}) \cosh (g ξ^{ξ^{'}}), \\ x^{2} = ξ^{2^{'}}, \\ (6.17) & x^{3} = ξ^{3^{'}} . \end{aligned}

The surfaces of constant

ξ^{0^{'}}

are the hyperplanes with

x^{0} / x^{1} = \tanh g ξ^{g^{'}}

sketched in Figure 6.4. Substitute expressions (6.17) into the Minkowski formula for the line element to find

\begin{aligned} d s^{2} & = η_{μ ν} d x^{μ} d x^{ν} \\ (6.18) & = - {(1 + g ξ^{1})}^{2} {(d ξ^{'})}^{2} + {(d ξ^{1})}^{2} + {(d ξ^{2})}^{2} + {(d ξ^{3})}^{2} . \end{aligned}

Figure 6.4.
Local coordinate system associated with an observer in hyperbolic motion (heavy black world line). The local coordinate system fails for

ξ^{1}

less than

- g^{- 1}

The coefficients of

d ξ^{μ^{'}} d ξ^{ν^{'}}

in this expansion are not the standard Lorentz metric components. The reason is clear. The

ξ^{μ^{'}}

do not form an inertial coordinate system. However, at the position of the observer,

ξ^{1^{'}} = 0

, the coefficients reduce to the standard form. Therefore these "local coordinates" approximate a Lorentz coordinate system in the immediate neighborhood of the observer.

EXERCISES

Exercise 6.6. CLOCK RATES VERSUS COORDINATE TIME IN ACCELERATED COORDINATES

Let a clock be attached to each grid point,

(ξ^{1}, ξ^{2}, ξ^{3^{'}}) =

constant, of the local coordinate system of an accelerated observer. Assume for simplicity that the observer is in hyperbolic motion. Use equation (6.18) to show that proper time as measured by a lattice clock differs from coordinate time at its lattice point:

d τ / d ξ^{0^{'}} = 1 + g ξ^{1^{'}}

(Of course, very near the observer, at

ξ^{1^{'}} ≪ g^{- 1}

, the discrepancy is negligible.)

Exercise 6.7. ACCELERATION OF LATTICE POINTS IN ACCELERATED COORDINATES

Let an accelerometer be attached to each grid point of the local coordinates of an observer in hyperbolic motion. Calculate the magnitude of the acceleration measured by the accelerometer at

(ξ^{1^{'}}, ξ^{2^{'}}, ξ^{3^{'}})

Exercise 6.8. OBSERVER WITH ROTATING TETRAD

An observer moving along an arbitrarily accelerated world line chooses not to Fermi-Walker transport his orthonormal tetrad. Instead, he allows it to rotate. The antisymmetric rotation tensor

Ω

that enters into his transport law

\begin{matrix} (6.19) & d e_{α^{'}} / d τ = - Ω \cdot e_{α^{'}} \end{matrix}

splits into a Fermi-Walker part plus a spatial rotation part:

\begin{matrix} (6.20) & Ω^{μ ν} = \underset{Ω_{(F W)}^{μ ν}}{\underset{⏟}{a^{μ} u^{ν} - a^{ν} u^{μ}}} + \underset{Ω_{(S R)}^{μ v}}{\underset{⏟}{u_{α} ω_{β} ϵ^{α β μ ν}}} \end{matrix}

ω =

a vector orthogonal to 4 -velocity

u

.
(a) The observer chooses his time basis vector to be

e_{0^{'}} = u

. Show that this choice is permitted by his transport law (6.19), (6.20).
(b) Show that

Ω_{(S R)}^{μ ν}

produces a rotation in the plane perpendicular to

u

and

ω

-i.e., that

\begin{matrix} (6.21) & Ω_{(S R)} \cdot u = 0, Ω_{(S R)} \cdot w = 0 \end{matrix}

e_{α^{''}}

. Show that the space vectors of his first tetrad rotate relative to those of his second tetrad with angular velocity vector equal to

ω

. Hint: At a moment when the tetrads coincide, show that (in three-dimensional notation, referring to the 3 -space orthogonal to the observer's world line):

\begin{matrix} (6.22) & d (e_{j^{'}} - e_{j^{''}}) / d τ = ω \times e_{j^{'}} \end{matrix}

(d) The observer uses the same prescription [equation (6.16)] to set up local coordinates based on his rotating tetrad as for his Fermi-Walker tetrad. Pick an event

Q

on the observer's world line, set

τ = 0

there, and choose the original inertial frame of prescription (6,16) so (1) it comoves with the accelerated observer at

Q, (2)

its origin is at

Q

, and (3) its axes coincide with the accelerated axes at

Q

. Show that these conditions translate into

\begin{matrix} (6.23) & z^{μ} (0) = 0, e_{α^{'}} (0) = e_{α^{'}} \end{matrix}

(e) Show that near

Q

, equations

(6.16)

for the rotating, accelerated coordinates reduce to:

\begin{aligned} (6.24) & x^{0} = ξ^{0^{'}} + a_{k^{'}} ξ^{k^{'}} ξ^{0^{'}} + O ({[ξ^{α^{'}}]}^{3}) \\ x^{j} = ξ^{j^{'}} + \frac{1}{2} a^{j} ξ^{0^{' 2}} + ϵ^{j k i} ω^{k} ξ^{ℓ^{'}} ξ^{0^{'}} + O ({[ξ^{α^{'}}]}^{3}) \end{aligned}

(f) A freely moving particle passes through the event

Q

with ordinary velocity

v

as measured in the inertial frame. By transforming its straight world line

x^{j} = v^{j} x^{0}

to the accelerated, rotating coordinates, show that its coordinate velocity and acceleration there are:

\begin{aligned} {(d ξ^{j^{'}} / d ξ^{0})}_{at 2} = v^{j}; \\ (6.25) & {(d^{2} ξ^{j} / d ξ^{0^{2} 2})}_{at 2} = \underset{an}{\underset{⏟}{- a^{j}}} - \underset{\begin{matrix} Coriolis \\ acceleration \end{matrix}}{\underset{⏟}{2 ϵ^{j k i} ω^{k} v^{l}}} + \underset{\begin{matrix} relativisitc \\ correction to \\ inertial acceleration \end{matrix}}{\underset{⏟}{2 v^{j} a^{k} v^{k}}} . \\ inertial \\ acceleration \end{aligned}

Exercise 6.9. THOMAS PRECESSION

Consider a spinning body (gyroscope, electron, ...) that accelerates because forces act at its center of mass. Such forces produce no torque; so they leave the body's intrinsic angularmomentum vector

S

unchanged, except for the unique rotation in the

u \land a

plane required to keep

S

orthogonal to the 4 -velocity

u

. Mathematically speaking, the body Fermi-Walker transports its angular momentum (no rotation in planes other than

u \land a

\begin{matrix} (6.26) & d S / d τ = (u \land a) \cdot S \end{matrix}

This transport law applies to a spinning electron that moves in a circular orbit of radius

r

around an atomic nucleus. As seen in the laboratory frame, the electron moves in the

x

y

-plane with constant angular velocity,

ω

. At time

t = 0

, the electron is at

x = r, y = 0

; and its spin (as treated classically) has components

S^{0} = 0, S^{x} = \frac{1}{\sqrt{2}} ℏ, S^{y} = 0, S^{z} = \frac{1}{2} ℏ

Calculate the subsequent behavior of the spin as a function of laboratory time,

S^{μ} (t)

. Answer:

\begin{aligned} S^{x} & = \frac{1}{\sqrt{2}} ℏ (\cos ω t \cos ω γ t + γ \sin ω t \sin ω γ t) \\ (6.27) & S^{y} & = \frac{1}{\sqrt{2}} ℏ (\sin ω t \cos ω γ t - γ \cos ω t \sin ω γ t) \\ S^{z} & = \frac{1}{2} ℏ; S^{0} = - \frac{1}{\sqrt{2}} ℏ v γ \sin ω γ t \\ v & = ω r; γ = {(1 - v^{2})}^{- 1 / 2} \end{aligned}

Rewrite the time-dependent spatial part of this as

\begin{matrix} (6.28) & S^{x} + i S^{y} = \frac{ℏ}{\sqrt{2}} [e^{- i (γ - 1) ω t} + i (1 - γ) \sin (ω γ t) e^{i ω t}] \end{matrix}

The first term rotates steadily in a retrograde direction with angular velocity

\begin{aligned} ω_{Thomas} & = (γ - 1) ω \\ (6.29) & \approx \frac{1}{2} v^{2} ω ≪ ω if v ≪ 1 . \end{aligned}

It is called the Thomas precession. The second term rotates in a righthanded manner for part of an orbit

(0 < ω γ t < π)

and in a lefthanded manner for the rest (

π < ω γ t < 2 π

). Averaged in time, it does nothing. Moreover, in an atom it is very small

(γ - 1 ≪ 1)

. It must be present, superimposed on the Thomas precession, in order to keep

\begin{matrix} (6.30) & s \cdot u = S \cdot u - S^{0} u^{0} = 0 \end{matrix}

and

\begin{matrix} (6.31) & S^{2} = S^{2} - {(S^{0})}^{2} = 3 ℏ^{2} / 4 = constant . \end{matrix}

It comes into play with righthanded rotation when

S \cdot u

is negative; it goes out of play when

S \cdot u = 0

; and it returns with lefthanded rotation when

S \cdot u

turns positive.

The Thomas precession can be understood, alternatively, as a spatial rotation that results from the combination of successive boosts in slightly different directions. [See, e.g., exercise 103 of Taylor and Wheeler (1966).] For an alternative derivation of the Thomas precession (6.29) from "spinor formalism," see §41.4.

CHAPTER

INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY

§7.1. ATTEMPTS TO INCORPORATE GRAVITY INTO SPECIAL RELATIVITY

The discussion of special relativity so far has consistently assumed an absence of gravitational fields. Why must gravity be ignored in special relativity? This chapter describes the difficulties that gravitational fields cause in the foundations of special relativity. After meeting these difficulties, one can appreciate fully the curved-spacetime methods that Einstein introduced to overcome them.

Start, then, with what one already knows about gravity, Newton's formulation of its laws:

\begin{aligned} (7.1) & d^{2} x^{i} / d t^{2} & = - \partial Φ / \partial x^{i}, \\ (7.2) & \nabla^{2} Φ & = 4 π G ρ . \end{aligned}

These equations cannot be incorporated as they stand into special relativity. The equation of motion (7.1) for a particle is in three-dimensional rather than four-dimensional form; it requires modification into a four-dimensional vector equation for

d^{2} x^{μ} / d τ^{2}

. Likewise, the field equation (7.2) is not Lorentz-invariant, since the appearance of a three-dimensional Laplacian operator instead of a four-dimensional d'Alembertian operator means that the potential

Φ

responds instantaneously to changes in the density

ρ

at arbitrarily large distances away. In brief, Newtonian gravitational fields propagate with infinite velocity.

One's first reaction to these problems might be to think that they are relatively straightforward to correct. Exercises at the end of this section study some relatively straightforward generalizations of these equations, in which the gravitational potential

Φ

is taken to be first a scalar, then a vector, and finally a symmetric tensor field. Each of these theories has significant shortcomings, and all fail to agree with observations. The best of them is the tensor theory (exercise 7.3, Box 7.1), which, however,

This chapter is entirely Track 2.

It depends on no preceding Track-2 material. It is not needed as preparation for any later chapter, but will be helpful in Chapter 18 (weak gravitational fields), and in Chapters 38 and 39 (experimental tests and other theories of gravity).

Newton's gravitational laws must be modified into four-dimensional, geometric form

All straightforward modifications are unsatisfactory

Best modification (tensor theory in flat spacetime) is internally inconsistent; when repaired, it becomes general relativity.
is internally inconsistent and admits no exact solutions. This difficulty has been attacked in recent times by Gupta (1954, 1957, 1962), Kraichnan (1955), Thirring (1961), Feynman (1963), Weinberg (1965), Deser (1970). They show how the flatspace tensor theory may be modified within the spirit of present-day relativistic field theory to overcome these inconsistencies. By this field-theory route (part 5 of Box 17.2), they arrive uniquely at standard 1915 general relativity. Only at this end point does one finally recognize, from the mathematical form of the equations, that what ostensibly started out as a flat-space theory of gravity is really Einstein's theory, with gravitation being a manifestation of the curvature of spacetime. This book follows Einstein's line of reasoning because it keeps the physics to the fore.

EXERCISES

EXERCISES ON FLAT-SPACETIME THEORIES OF GRAVITY

The following three exercises provide a solid challenge. Happily, all three require similar techniques, and a solution to the most difficult one (exercise 7.3) is presented in Box 7.1. Therefore, it is reasonable to proceed as follows. (a) Work either exercise 7.1 (scalar theory of gravity) or 7.2 (vector theory of gravity), skimming exercise 7.3 and Box 7.1 (tensor theory of gravity) for outline and method, not for detail, whenever difficulties arise. (b) Become familiar with the results of the other exercise ( 7.2 or 7.1 ) by discussing it with someone who has worked it in detail. (c) Read in detail the solution to exercise 7.3 as presented in Box 7.1, and compare with the computed results for the other two theories. (d) Develop computational power by checking some detailed computations from Box 7.1.

Exercise 7.1. SCALAR GRAVITATIONAL FIELD,

Φ

A. Consider the variational principle

δ I = 0

, where

\begin{matrix} (7.3) & I = - m \int e^{ϕ} {(- η_{α β} \frac{d z^{α}}{d λ} \frac{d z^{β}}{d λ})}^{1 / 2} d λ \end{matrix}

Here

m =

(rest mass) and

z^{α} (λ) =

(parametrized world line) for a test particle in the scalar gravitational field

Φ

. By varying the particle's world line, derive differential equations governing the particle's motion. Write them using the particle's proper time as the path parameter,

d τ = {(- η_{α β} \frac{d z^{α}}{d λ} \frac{d z^{β}}{d λ})}^{1 / 2} d λ

so that

u^{α} = d z^{α} / d τ

satisfies

u^{α} u^{β} η_{α β} = - 1

.
B. Obtain the field equation for

Φ (x)

implied by the variational principle

δ I = 0

, where

I = \int E d^{4} x

and

\begin{matrix} (7.4) & E = - \frac{1}{8 π G} η^{α β} \frac{\partial Φ}{\partial x^{α}} \frac{\partial Φ}{\partial x^{β}} - \int m e^{φ} δ^{4} [x - z (τ)] d τ \end{matrix}

Show that the second term here gives the same integral as that studied in part A (equation 7.3).

Discussion: The field equations obtained describe how a single particle of mass

m

generates the scalar field. If many particles are present, one includes in

E

a term

- \int m e^{ϕ} δ^{4} [x - z (τ)] d τ

for each particle.
C. Solve the field equation of part B, assuming a single source particle at rest. Also assume that

e^{ϕ} = 1

is an adequate approximation in the neighborhood of the particle. Then check this assumption from your solution; i.e., what value does it assign to

e^{ϕ}

at the surface of the earth? (Units with

c = 1

are used throughout; one may also set

G = 1

, if one wishes.)
D. Now treat the static, spherically symmetric field

Φ

from part C as the field of the sun acting as a given external field in the variational principle of part A , and study the motion of a planet determined by this variational principle. Constants of motion are available from the spherical symmetry and time-independence of the integrand. Use spherical coordinates and assume motion in a plane. Derive a formula for the perihelion precession of a planet.
E. Pass to the limit of a zero rest-mass particle in the equations of motion of part A. Do this by using a parameter

λ

different from proper time, so chosen that

k^{μ} = d x^{μ} / d λ

is the energy-momentum vector, and by taking the limit

m ⟶ 0

with

k^{0} = γ m = E

remaining finite (so

u^{0} = γ ⟶ \infty

). Use these equations to show that the quantities

q^{μ} = k^{μ} e^{ϕ}

are constants of motion, and from this deduce that there is no bending of light by the sun in this scalar theory.

Exercise 7.2. VECTOR GRAVITATIONAL FIELD, $Φ_{μ}$

A. Verify that the variational principle

δ I = 0

gives Maxwell's equations by varying

A_{μ}

, and the Lorentz force law by varying

z^{μ} (τ)

, when

\begin{matrix} (7.5) & I = \frac{- 1}{16 π} \int F_{μ ν} F^{μ ν} d^{4} x + \frac{1}{2} m \int \frac{d z^{μ}}{d τ} \frac{d z_{μ}}{d τ} d τ + e \int \frac{d z^{μ}}{d τ} A_{μ} (z) d τ \end{matrix}

Here

F_{μ ν}

is an abbreviation for

A_{v, μ} - A_{μ, ν}

. Hint: to vary

A_{μ} (x)

, rewrite the last term as a spacetime integral by introducing a delta function

δ^{4} [x - z (τ)]

as in exercise 7.1, parts A and B.
B. Define, by analogy to the above, a vector gravitational field

Φ_{μ}

with

G_{μ ν} \equiv Φ_{ν, μ} - Φ_{μ, ν}

using a variational principle with

\begin{matrix} (7.6) & I = + \frac{1}{16 π G} \int G_{μ ν} G^{μ v} d^{4} x + \frac{1}{2} m \int \frac{d z^{μ}}{d τ} \frac{d z_{μ}}{d τ} d τ + m \int Φ_{μ} \frac{d z^{μ}}{d τ} d τ \end{matrix}

(Note: if many particles are present, one must augment

I

by terms

\frac{1}{2} m \int (d z^{μ} / d τ) (d z_{μ} / d τ) d τ +

m \int Φ_{μ} (d z^{μ} / d τ) d τ

for each particle.) Find the "Coulomb" law in this theory, and verify that the coefficients of the terms in the variational principle have been chosen reasonably.
C. Compute the perihelion precession in this theory.
D. Compute the bending of light in this theory (i.e., scattering of a highly relativistic particle

u^{0} = γ ⟶ \infty

), as it passes by the sun, because of the sun's

Φ_{μ}

field.
E. Obtain a formula for the total field energy corresponding to the Lagrangian implicit in part B. Use the standard method of Hamiltonian mechanics, with

I_{field} = \frac{1}{16 π G} \int G_{μ ν} G^{μ v} d^{4} x \equiv \int E d^{4} x

L

is the Lagrangian density and

L \equiv \int E d^{3} x

is the Lagrangian. The corresponding Hamiltonian density (

\equiv

energy density) is

K = \sum_{μ} Φ_{μ, 0} \frac{\partial L}{\partial Φ_{μ, 0}} - E

Show that vector gravitational waves carry negative energy.
Exercise 7.3. SYMMETRIC TENSOR GRAVITATIONAL FIELD,

h_{μ ν} = h_{ν μ}

Here the action principle is, as for the vector field,

δ I = 0

, with

I = I_{field} + I_{particle} +

I_{interaction} . I_{particle}

is the same as for the vector field:

\begin{matrix} (7.7) & I_{particle} = \frac{1}{2} m \int \frac{d z^{μ}}{d τ} \frac{d z_{μ}}{d τ} d τ \end{matrix}

However,

I_{field}

and

I_{interaction}

are different:

\begin{matrix} (7.8b) & \begin{aligned} I_{field} & = \int E_{f} d^{4} x, \\ E_{f} & = \frac{- 1}{32 π G} (\frac{1}{2} h_{ν β, α} {\bar{h}}^{ν β, α} - {\bar{h}}_{μ α}, α \\ h^{μ β} \\ , β \end{aligned}) [\begin{array}{l} Note that \\ one h here \\ is not an \bar{h} \end{array}], $ \end{matrix}

with

\begin{aligned} (7.8c) & {\bar{h}}_{μ ν} & \equiv h_{μ ν} - \frac{1}{2} η_{μ ν} h_{σ}^{σ} \\ (7.9) & I_{interaction} & = \frac{1}{2} \int h_{μ ν} T^{μ ν} d^{4} x . \end{aligned}

Here

T^{μ ν}

is the stress-energy tensor for all nongravitational fields and matter present. For a system of point particles (used throughout this exercise),

\begin{matrix} (7.10) & T^{μ ν} (x) = \int m \frac{d z^{μ}}{d τ} \frac{d z^{ν}}{d τ} δ^{4} [x - z (τ)] d τ \end{matrix}

A. Obtain the equations of motion of a particle by varying

z^{μ} (τ)

δ (I_{particle} +

I_{interaction}) = 0

. Express your result in terms of the "gravitational force field"

\begin{matrix} (7.11) & Γ_{ν α β} = \frac{1}{2} (h_{ν α, β} + h_{ν β, α} - h_{α β, ν}) \end{matrix}

derived from the tensor gravitational potentials

h_{μ ν} = h_{ν μ}

.
B. Obtain the field equations from

δ (I_{field} + I_{interaction}) = 0

; express them in terms of

\begin{matrix} (7.12) & - H^{μ α ν β} \equiv {\bar{h}}^{μ ν} η^{α β} + {\bar{h}}^{α β} η^{μ ν} - {\bar{h}}^{α ν} η^{μ β} - {\bar{h}}^{μ β} η^{α ν} \end{matrix}

Discuss gauge invariance, and the condition

{\bar{h}}^{μ α}_{, α} = 0

.
C. Find the tensor gravitational potentials

h_{μ ν}^{, α}

due to the sun (treated as a point mass).
D. Compute the perihelion precession.
E. Compute the bending of light.
F. Consider a gravitational wave

\begin{matrix} (7.13) & {\bar{h}}^{μ ν} = A^{μ ν} \exp (i k_{α} x^{α}) \end{matrix}

What conditions are imposed by the field equations? By the gauge condition

\begin{matrix} (7.14) & {\bar{h}}^{μ α}_{, α} = 0 ? \end{matrix}

Show that, by further gauge transformations

\begin{matrix} (7.15) & h_{μ ν} ⟶ h_{μ ν} + ξ_{μ, ν} + ξ_{v, μ} \end{matrix}

that preserve the

{\bar{h}}^{μ α}_{, α} = 0

restrictions, further conditions

\begin{matrix} (7.16) & u_{α} {\bar{h}}^{α μ} = 0, {\bar{h}}_{α}^{α} = 0 \end{matrix}

can be imposed, where

u^{α}

is a fixed, timelike vector. It is sufficient to consider the case, obtained by a suitable choice of reference frame, where

u^{α} = (1; 0, 0, 0)

and

k^{α} = (ω; 0, 0, ω)

.
G. From the Hamiltonian density

\begin{matrix} (7.17) & H \equiv {\dot{h}}_{μ ν} (\partial L / \partial {\dot{h}}_{μ ν}) - L \end{matrix}

for the field, show that the energy density of the waves considered in part F is positive.
H. Compute

T^{μ ν},^{'}

, for the stress-energy tensor of particles

T^{μ ν}

that appears in the action integral

I

. Does

T^{μ ν}, v

vanish (e.g., for the earth in orbit around the sun)? Why? Show that the coupled equations for fields and particles obtained from

δ I = 0

have no solutions.

Box 7.1 AN ATTEMPT TO DESCRIBE GRAVITY BY A SYMMETRIC TENSOR FIELD IN FLAT SPACETIME [Solution to exercise 7.3]

Attempts to describe gravity within the framework of special relativity would naturally begin by considering the gravitational field to be a scalar (exercise 7.1) as it is in Newtonian theory, or a vector (exercise 7.2) by analogy to electromagnetism. Only after these are found to be deficient (e.g., no bending of light in either theory; negative-energy waves in the vector theory) would one face the computational complexities of a symmetric tensor gravitational potential,

h_{μ ν} = h_{ν μ}

, which has more indices.

The foundations of the most satisfactory of all tensor theories of gravity in flat spacetime are laid out at the beginning of exercise 7.3. The choice of the Lagrangian made there (equations 7.8) is dictated by the demand that

h_{μ ν}

be a "Lorentz covariant, massless, spin-two field." The meaning of this demand, and the techniques of special relativity required to translate it into a set of field equations, are customarily found in books on elementary particle physics or quantum field theory; see, e.g., Wentzel (1949), Feynman (1963), or Gasiorowicz (1966). Fierz and Pauli (1939) were the first to write down this Lagrangian and investigate the resulting theory. The conclusions of the theory are spelled out here in the form of a solution to exercise 7.3.

A. Equation of Motion for a Test Particle (exercise 7.3A)

Carry out the integration in equation (7.9), using the particle stress-energy tensor of equation (7.10), to find

\begin{matrix} (1) & I_{p + i} \equiv I_{particle} + I_{interaction} = \frac{1}{2} m \int (η_{μ ν} + h_{μ ν}) {\dot{z}}^{μ} {\dot{z}}^{ν} d τ \end{matrix}

where

{\dot{z}}^{μ} \equiv d z^{μ} / d τ

Then compute

δ I_{p + i}

, and find that the coefficient of the arbitrary variation in path

δ z^{μ}

vanishes if and only if

(d / d τ) [(η_{μ ν} + h_{μ ν}) {\dot{z}}^{ν}] - \frac{1}{2} h_{α β, μ} {\dot{z}}^{α} {\dot{z}}^{β} = 0

Rewrite this equation of motion in the form

\begin{matrix} (2) & (η_{μ ν} + h_{μ ν}) {\ddot{z}}^{ν} + Γ_{μ α β} {\dot{z}}^{α} {\dot{z}}^{β} = 0 \end{matrix}

where

Γ_{μ α β}

is defined in equation (7.11).

$B_{1}$ . Field Equations (exercise 7.3B)

Use

I_{field}

and

I_{interaction}

in the forms given in equations (7.8) and (7.9); but for the quickest and least messy derivation, do not use the standard Euler-Lagrange equations. Instead, compute directly the first-order change

δ L_{f}

produced by a small

Box 7.1 (continued)

variation

δ h_{α β}

of the field. For the second term of

E_{ρ}

, it is clear (by relabeling dummy indices as needed) that varying each factor gives the same result, so the two terms from the product rule combine:

δ ({\bar{h}}_{μ α^{'}} {\bar{h}}^{μ β}_{, β}) = 2 {\bar{h}}^{μ β}_{, β} δ {\bar{h}}_{μ α^{'}, α} .

A similar result holds for the first term of

E_{f}

, in view of the identity

a_{μ ν} {\bar{b}}^{μ ν} = {\bar{a}}_{μ ν} b^{μ ν}

, which holds for the "bar" operation of equations (7.8); each side here is just

a_{μ ν} b^{μ ν}

- \frac{1}{2} a^{μ}_{μ} b^{ν}_{p}

. Consequently,

\begin{matrix} (3) & - (32 π G) δ Q_{f} = {\bar{h}}^{ν β, α} δ h_{ν β, α} - 2 {\bar{h}}^{μ β}_{, β} δ {\bar{h}}_{μ α^{'}}^{α} . \end{matrix}

Next use this expression in

δ I_{field}

; and, by an integration by parts, remove the derivatives from

δ h_{μ ν}

, giving

δ I_{field} = (32 π G)^{- 1} \int [{\bar{h}}^{ν β, α}_{, α} δ h_{ν β} - 2 {\bar{h}}_{, β}^{μ β},_{, α} δ {\bar{h}}_{μ α}] d^{4} x

To find the coefficient of

δ h_{μ ν}

in this expression, write (from equation 7.8c)

δ {\bar{h}}_{α β} = (δ_{α}^{μ} δ_{ν β} - \frac{1}{2} η_{α β} η^{μ ν}) δ h_{μ ν};

and then rearrange and relabel dummy (summation) indices to obtain

δ I_{field} = (32 π G)^{- 1} \int [{\bar{h}}^{μ β, α}_{, α} δ h_{ν β} - 2 {\bar{h}}^{μ β}_{, β}^{, α} δ {\bar{h}}_{μ α}] d^{4} x

By combining this with

δ I_{interaction} = \frac{1}{2} T^{μ ν} δ h_{μ ν} d^{4} x

, and by using the symmetry

δ h_{μ ν} = δ h_{ν μ}

, obtain

\begin{matrix} (4) & - {\bar{h}}^{μ ν}_{, α^{α}} - η^{μ ν} {\bar{h}}^{α β}_{, α β} + {\bar{h}}^{μ α}_{, α^{, ν}} + {\bar{h}}^{ν α}_{, α^{, μ}} = 16 π G T^{μ ν} . \end{matrix}

The definition made in equation (7.12) allows this to be rewritten as

\begin{matrix} (') & H^{μ α ν β}_{, α β} = 16 π G T^{μ ν} . \end{matrix}

$B_{2}$ . Gauge Invariance (exercise 7.3B, continued)

The symmetries,

H^{μ α ν β} = H^{[μ α] [ν β]} = H^{ν β μ α},

H^{μ α ν β}

imply an identity

H^{μ α ν β}_{, α β ν} = H^{μ α [ν β 1}_{, α (β ν)} \equiv 0

analogous to

F^{μ ν}_{, v μ} \equiv 0

in electromagnetism.
Thus

T^{μ ν}_{, ν} = 0

is required of the sources, just as is

J^{μ}_{, μ} = 0

in electromagnetism (exercise 3.16). These identities make the field equations (4') too weak to fix

h_{μ ν}

completely. In particular, by direct substitution in equations (4), one verifies that to any solution one can add a gauge field

\begin{aligned} (5) & h_{μ ν}^{(gauge)} = ξ_{μ, ν} + ξ_{ν, μ}, \\ {\bar{h}}_{μ ν}^{(gauge)} = ξ_{μ, ν} + ξ_{ν, μ} - η_{μ ν} ξ^{α}_{, α}, \end{aligned}

without changing

T^{μ ν}

.
Let

ξ_{μ}

vanish outside some finite spacetime volume, but be otherwise arbitrary. Then

h_{μ ν}

and

h_{μ ν} = h_{μ ν} + h_{μ ν}^{(gauge)}

both satisfy the source equation (4) for the same source

T^{μ ν}

and the same boundary conditions at infinity. We therefore expect them to be physically equivalent.
By a specialization of the gauge analogous to the "Lorentz" specialization

A^{α}_{, α} = 0

of electromagnetism (equation 3.58a; exercise 3.17), one imposes the condition

\begin{matrix} (6) & {\bar{h}}^{μ α}_{, α} = 0 . \end{matrix}

This reduces the field equations (4) to the simple d'Alembertian form

\begin{matrix} (7) & ◻ {\bar{h}}^{μ ν} \equiv {\bar{h}}^{μ ν}_{, α}^{α} = - 16 π T^{μ ν} \end{matrix}

(see exercise 18.2). Here and henceforth we set

G = 1

("geometrized units").

C. Field of a Point Mass (exercise 7.3C)

For a static source, the wave equation (7) reduces to a Laplace equation

\nabla^{2} {\bar{h}}_{μ ν} = - 16 π T_{μ ν}

The stress-energy tensor for a static point mass (equation 7.10) is

T^{00} = M δ^{3} (x)

and

T^{μ k} = 0

. Put this into the Laplace equation, solve for

{\bar{h}}_{μ ν}

, and use equation (7.8c) to obtain

h_{μ ν}

. The result is:

\begin{matrix} (8) & h_{00} = 2 M / r; h_{0 k} = 0; h_{i k} = δ_{i k} (2 M / r) \end{matrix}

(see equation 18.15a).

D. Perihelion Precession (exercise 7.3D)

Direct substitution of the potential (8) into the equations of motion (2) is tedious and not very instructive. Variational principles are popular in mechanics because they simplify such calculations. Return to the basic variational principle

δ I_{p + i} = 0

(equation 1), and insert the potential (8) for the sun. Convert to spherical coordinates so oriented that the orbit lies in the equatorial

(θ = π / 2)

plane:

\begin{aligned} (9) & I_{p + i} = \int L d τ \\ (10) & L = \frac{1}{2} m [- (1 - 2 M r^{- 1}) {\dot{t}}^{2} + (1 + 2 M r^{- 1}) ({\dot{r}}^{2} + r^{2} ϕ^{2})] \end{aligned}

Box 7.1 (continued)

From the absence of explicit

t

ϕ

-, and

τ

-dependence in

L

, infer three constants of motion: the canonical momenta

P_{t} \equiv - m γ = \partial L / \partial \dot{i}

(this defines

γ

) and

P_{ϕ} \equiv m α = \partial L / \partial \dot{ϕ}

(this defines

α

); and the Hamiltonian

H = {\dot{x}}^{μ} (\partial L / \partial {\dot{x}}^{μ}) - L,

which can be set equal to

- \frac{1}{2} m

by appropriate normalization of the path parameter

τ

. From these constants of the motion, derive an orbit equation as follows: (1) calculate

H = - \frac{1}{2} m

in terms of

r, \dot{r}, \dot{ϕ}

, and

\dot{i}

; (2) eliminate

\dot{i}

and

\dot{ϕ}

in favor of the constants

γ

and

α

; (3) as in Newtonian orbit problems, define

u = M / r

, and write

\frac{d u}{d \dot{ϕ}} = \frac{\dot{u}}{\dot{ϕ}} = - \frac{M \dot{r}}{r^{2} \dot{ϕ}} = - \frac{M}{α} (1 + 2 u) \dot{r};

(4) in

H

, eliminate

\dot{r}

in favor of

d u / d \dot{ϕ}

via the above equation, and eliminate

r

in favor of

u

; (5) solve for

d u / d ϕ

. The result is

\begin{matrix} (11) & {(\frac{d u}{d ϕ})}^{2} + u^{2} = (γ^{2} - 1 + 2 u) \frac{M^{2}}{α^{2}} [\frac{1 + 2 u}{1 - 2 u}] \end{matrix}

Neglecting cubic and higher powers of

u = G M / c^{2} r \sim (1 - γ^{2})

in this equation, derive the perihelion shift. (For details of method, see exercise 40.4, with the

γ

and

α

of this box renamed

\tilde{E}

and

\tilde{L}

, and with the

γ

and

β

of that exercise set equal to 1 and 0 .) The resulting shift per orbit is

\begin{matrix} (12) & Δ ϕ = 8 π M / r_{0} + O ({[M / r_{0}]}^{2}) . \end{matrix}

This is

\frac{4}{3}

the prediction of general relativity; and it disagrees with the observations on Mercury (see Box 40.3).

E. Bending of Light (exercise 7.3E)

The deflection angle for light passing the sun is, on dimensional grounds, a small quantity,

Δ ϕ \sim M_{⊙} / R_{⊙} \sim 10^{- 6}

; from the outset, one makes approximations based on this smallness. A diagram of the photon trajectory, set in the

x, z

-plane, shows that, for initial motion parallel to the

z

-axis, the deflection angle can be expressed in terms of the final momentum as

Δ ϕ = p_{x} / p_{z}

. Compute the final

p_{x}

by an integral along the trajectory,

p_{x} = \int_{- \infty}^{+ \infty} (d p_{x} / d z) d z

treating

p_{z}

as essentially constant. This computation requires generalization of the equation of motion (2) to the case of zero rest mass. To handle the limit

m ⟶ 0

, introduce a new parameter

λ = τ / m

; then

p^{μ} = m (d z^{μ} / d τ) = d z^{μ} / d λ

. Also define

p_{μ} = (η_{μ ν} + h_{μ ν}) p^{ν}

, since this quantity appears simply in equation (2) and agrees with

p_{μ}

in the limit

r ⟶ \infty

, where one will need to evaluate it. Then equation (2) reads, for any

m

, including

m = 0

\frac{d P_{μ}}{d λ} = \frac{1}{2} h_{α β, μ} p^{α} p^{β}

On the righthand side here, since

h_{α β, μ}

is small, a crude approximation to

p^{μ}

suffices;

p^{1} = p^{2} = 0, p^{0} = p^{3} = d z / d λ = ω =

constant. Thus,

\frac{d P_{1}}{d λ} = \frac{1}{2} {(h_{00} + 2 h_{03} + h_{33})}_{1} ω^{2}

and

\frac{1}{p_{3}} \frac{d P_{1}}{d z} = \frac{1}{2} {(h_{00} + 2 h_{03} + h_{33})}_{, 1}

For the sun,

h_{00} = h_{33} = 2 M / r

, and

h_{03} = 0

(equation 8 ), so

\begin{matrix} (13) & Δ ϕ = - {(\frac{p_{1}}{p_{3}})}_{final} = - {(\frac{P_{1}}{p_{3}})}_{final} = \int_{- \infty}^{\infty} \frac{2 M ℓ d z}{{(ℓ^{2} + z^{2})}^{3 / 2}} = \frac{2 M}{ℓ} \int_{- \infty}^{\infty} \frac{d ζ}{{(1 + ζ^{2})}^{3 / 2}} = \frac{4 M}{ℓ} \end{matrix}

For light grazing the sun,

ℓ = R_{⊙}

, this gives

Δ ϕ = 4 M_{⊙} / R_{⊙}

radians

= 1^{''} .75

, which is also the prediction of general relativity, and is consistent with the observations (see Box 40.1).

F. Gravitational Waves (exercise 7.3F)

The field equations (4) and gauge properties (5) of the present flat-spacetime theory are identical to those of Einstein's "linearized theory." Thus, the treatment of gravitational waves using linearized theory, which is presented in

$ $ 18.2, 35.3

, and 35.4, applies here.

G. Positive Energy of the Waves (exercise 7.3G)

Computing a general formula for

K

from equation (7.17) is tedious, but it is sufficient to consider only the special case of a plane wave (equation 7.13)-or better still,

Box 7.1 (continued)
a plane wave with only

h_{12} = h_{21} = f (z - t)

. Any gravitational wave can be constructed as a superposition of such plane waves. First compute the Langrangian for this case. According to equation (7.8), it reads

E_{f} = (32 π)^{- 1} [{(h_{12, 0})}^{2} - {(h_{12, 3})}^{2}] .

Now the full content of the formula (7.17) defining

K

is precisely the following: start from the Lagrangian; keep all terms that are quadratic in time derivatives; omit all terms that are linear in time derivatives; and reverse the sign of terms that contain no time derivatives. The result is

\begin{matrix} (14) & K = (32 π)^{- 1} [{(h_{12, 0})}^{2} + {(h_{12, 3})}^{2}], \end{matrix}

which is positive.

H. Self-Inconsistency of the Theory (exercise 7.3H)

From equation (7.10), find

T^{μ ν}_{, ν} = m \int \frac{d z^{μ}}{d τ} \frac{d z^{ν}}{d τ} \frac{\partial}{\partial x^{ν}} δ^{4} [x - z (τ)] d τ .

But

δ^{4} (x - z)

depends only on the difference

x^{μ} - z^{μ}

, so

- \partial / \partial z^{ν}

can replace

\partial / \partial x^{ν}

when acting on the

δ

-function. Noting that

\frac{d z^{v}}{d τ} \frac{\partial}{\partial z^{ν}} δ^{4} [x - z (τ)] = \frac{d}{d τ} δ^{4} [x - z (τ)],

rewrite

T^{μ ν}_{, v}

T^{μ ν}_{, ν} = - m \int {\dot{z}}^{μ} (d / d τ) δ^{4} [x - z (τ)] d τ = + m \int {\ddot{z}}^{μ} δ^{4} [x - z (τ)] d τ

(The last step is obtained by an integration by parts.) Thus

T^{μ ν}_{, ν} = 0

holds if and only if

{\ddot{z}}^{μ} = 0

. But

{\ddot{z}}^{μ} = 0

means that the gravitational fields have no effect on the motion of the particle. But this contradicts the equation of motion (2), which follows from the theory's variational principle. Thus, this tensor theory of gravity is inconsistent. [Stated briefly, equation (4) requires

T^{μ v}_{, ν} = 0

, while equation (2) excludes it.]

The fact that, in this theory, gravitating bodies cannot be affected by gravity, also holds for bodies made of arbitrary stress-energy (e.g., rubber balls or the Earth). Since all bodies gravitate, since the field equations imply

T^{μ ν}_{, ν} = 0

, and since this "equation of motion for stress-energy" implies conservation of a body's total 4-momentum

P^{μ} = \int T^{μ 0} d^{3} x

, no body can be accelerated by gravity. The Earth cannot be attracted by the sun; it must fly off into interstellar space!
Straightforward steps to repair this inconsistency in the theory lead inexorably to general relativity (see Box 17.2 part 5). Having adopted general relativity as the theory of gravity, one can then use the present flat-spacetime theory as an approximation to it ("Linearized general relativity"; treated in Chapters 18, 19, and 35 ; see especially discussion at end of §18.3).

§7.2. GRAVITATIONAL RED SHIFT DERIVED FROM ENERGY CONSERVATION

Einstein argued against the existence of any ideal, straight-line reference system such as is assumed in Newtonian theory. He emphasized that nothing in a natural state of motion, not even a photon, could ever give evidence for the existence or location of such ideal straight lines.

That a photon must be affected by a gravitational field Einstein (1911) showed from the law of conservation of energy, applied in the context of Newtonian gravitation theory. Let a particle of rest mass

m

start from rest in a gravitational field

g

at point

C

and fall freely for a distance

h

to point

G

. It gains kinetic energy

m g h

. Its total energy, including rest mass, becomes

\begin{matrix} (7.18) & m + m g h . \end{matrix}

Now let the particle undergo an annihilation at

B

, converting its total rest mass plus kinetic energy into a photon of the same total energy. Let this photon travel upward in the gravitational field to

a

. If it does not interact with gravity, it will have its original energy on arrival at

a

. At this point it could be converted by a suitable apparatus into another particle of rest mass

m

(which could then repeat the whole process) plus an excess energy

m g h

that costs nothing to produce. To avoid this contradiction of the principal of conservation of energy, which can also be stated in purely classical terms, Einstein saw that the photon must suffer a red shift. The energy of the photon must decrease just as that of a particle does when it climbs out of the gravitational field. The photon energy at the top and the bottom of its path through the gravitational field must therefore be related by

\begin{matrix} (7.19) & E_{bottom} = E_{top} (1 + g h) = E_{top} (1 + g_{conv} h / c^{2}) \end{matrix}

The drop in energy because of work done against gravitation implies a drop in frequency and an increase in wavelength (red shift; traditionally stated in terms of a red shift parameter,

z = Δ λ / λ

); thus,

\begin{matrix} (7.20) & 1 + z = \frac{λ_{top}}{λ_{bottom}} = \frac{h ν_{bottom}}{h v_{top}} = \frac{E_{bottom}}{E_{top}} = 1 + g h . \end{matrix}

The redshift predicted by this formula has been verified to 1 percent by Pound and Snider (1964, 1965), refining an experiment by Pound and Rebka (1960).

§7.3. GRAVITATIONAL REDSHIFT IMPLIES SPACETIME IS CURVED

An argument by Schild

(1960, 1962, 1967)

yields an important conclusion: the existence of the gravitational redshift shows that a consistent theory of gravity cannot be constructed within the framework of special relativity.

Gravitational redshift derived from energy considerations

Figure 7.1.
Successive pulses of light rising from height

z_{1}

, to height

z_{2} = z_{1} + h

against the gravitational field of the earth. The paths

γ_{1}

, and

γ_{2}

must be exactly congruent, whether sloped at

45^{\circ}

(left) or having variable slope (right).

Assume gravity is described by an (unspecified) field in flat spacetime . . .

Whereas Einstein's argument (last section) was formulated in Newtonian theory, Schild's is formulated in special relativity. It analyzes gravitational redshift experiments in the field of the Earth, using a global Lorentz frame tied to the Earth's center. It makes no demand that free particles initially at rest remain at rest in this global Lorentz frame (except far from the Earth, where gravity is negligible). On the contrary, it demands that free particles be accelerated relative to the Lorentz frame by the Earth's gravitational field. It is indifferent to the mathematical nature of that field (scalar, vector, tensor, ...), but it does insist that the gravitational accelerations agree with experiment. And, of course, it demands that proper lengths and times be governed by the metric of special relativity.

Schild's argument proceeds as follows. Consider one observer at rest on the Earth's surface at height

z_{1}

, and a second above the Earth's surface at height

z_{2} = z_{1} + h

(Figure 7.1). The observers may verify that they are at rest relative to each other and relative to the Earth's Lorentz frame by, for instance, radar ranging to free particles that are at rest in the Earth's frame far outside its gravitational field. The bottom experimenter then emits an electromagnetic signal of a fixed standard frequency

ω_{b}

which is received by the observer on top. For definiteness, let the signal be a pulse exactly

N

cycles long. Then the interval of time*

δ τ_{bot}

required for the emission of the pulse is given by

2 π N = ω_{b} δ τ_{bot}

. The observer at the top is then to receive these same

N

cycles of the electromagnetic wave pulse and measure the time interval*

δ τ_{top}

required. By the definition of "frequency," it satisfies

2 π N =

ω_{t} δ τ_{top}

. The redshift effect, established by experiment (for us) or by energy conservation (for Einstein), shows

ω_{t} < ω_{b}

; consequently the time intervals are different,

δ τ_{top} > δ τ_{bot}

. Transfer this information to the special-relativity spacetime diagram of the experiment (Figure 7.1). The waves are light rays; so one might show them as traveling along

45^{\circ}

null lines in the spacetime diagram (Figure 7.1,A). In this

simplified but slightly inadequate form of the argument, one reaches a contradiction by noticing that here one has drawn a parallelogram in Minkowski spacetime in which two of the sides are unequal,

τ_{top} > τ_{bot}

, whereas a parallelogram in flat Minkowski spacetime cannot have opposite sides unequal. One concludes that special relativity cannot be valid over any sufficiently extended region. Globally, spacetime, as probed by the tracks of light rays and test particles, departs from flatness ("curvature"; Parts III and IV of this book), despite the fine fit that Lorentz-Minkowski flatness gives to physics locally.

Figure 7.1,B, repairs an oversimplification in this argument by recognizing that the propagation of light will be influenced by the gravitational field. Therefore photons might not follow straight lines in the diagram. Consequently, the world lines

γ_{1}

and

γ_{2}

of successive pulses are curves. However, the gravitational field is static and the experimenters do not move. Therefore nothing in the experimental setup changes with time. Whatever the path

γ_{1}

, the path

γ_{2}

must be a congruent path of exactly the same shape, merely translated in time. On the basis of this congruence and the fact that the observers are moving on parallel world lines, one would again conclude, if flat Minkowski geometry were valid, that

τ_{bot} = τ_{top}

, thus contradicting the observed redshift experiment. The experimenters do not need to understand the propagation of light in a gravitational field. They need only use their radar apparatus to verify the fact that they are at rest relative to each other and relative to the source of the gravitational field. They know that, whatever influence the gravitational field has on their radar apparatus, it will not be a time-dependent influence. Moreover, they do not have to know how to compute their separation in order to verify that the separation remains constant. They only need to verify that the round-trip time required for radar pulses to go out to each other and back is the same every time they measure it.

Schild's redshift argument does not reveal what kind of curvature must exist, or whether the curvature exists in the neighborhood of the observational equipment or some distance away from it. It does say, however, quite unambigously, that the flat spacetime of special relativity is inadequate to describe the situation, and it should therefore motivate one to undertake the mathematical analysis of curvature in Part III.

§7.4. GRAVITATIONAL REDSHIFT AS EVIDENCE FOR THE PRINCIPLE OF EQUIVALENCE

Einstein

(1908, 1911)

elevated the idea of the universality of gravitational interactions to the status of a fundamental principle of equivalence, that all effects of a uniform gravitational field are identical to the effects of a uniform acceleration of the coordinate system. This principle generalized a result of Newtonian gravitation theory, in which a uniform acceleration of the coordinate system in equation (7.1) gives rises to a

This assumption is incompatible with gravitational redshift

Conclusion: spacetime is curved

Principle of equivalence: a uniform gravitational field is indistinguishable from a uniform acceleration of a reference frame

Gravitational redshift derived from equivalence principle

Equivalence principle implies nonmeshing of local Lorentz frames near Earth (spacetime curvature!)
supplementary uniform gravitational field. However, the Newtonian theory only gives this result for particle mechanics. Einstein's principle of equivalence asserts that a similar correspondence will hold for all the laws of physics, including Maxwell's equations for the electromagnetic field.

The rules of the game - the "scientific method"-require that experimental support be sought for any new theory or principle, and Einstein could treat the gravitational redshift as the equivalent of experimental confirmation of his principle of equivalence. There are two steps in such a confirmation: first, the theory or principle must predict an effect (the next paragraph describes how the equivalence principle implies the redshift); second, the predicted effect must be observed. With the Pound-RebkaSnider experiments, one is in much better shape today than Einstein was for this second step. Einstein himself had to rely on the experiments supporting the general concept of energy conservation, plus the necessity of a redshift to preserve energy conservation, as a substitute for direct experimental confirmation.
The existence of the gravitational redshift can be deduced from the equivalence principle by considering two experimenters in a rocket ship that maintains a constant acceleration

g

. Let the distance between the two observers be

h

in the direction of the acceleration. Suppose for definiteness that the rocket ship was at rest in some inertial coordinate system when the bottom observer sent off a photon. It will require time

t = h

for the photon to reach the upper observer. In that time the top observer acquires a velocity

v = g t = g h

. He will therefore detect the photon and observe a Doppler redshift

z = v = g h

. The results here are therefore identical to equation (7.20). The principle of equivalence of course requires that, if this redshift is observed in an experiment performed under conditions of uniform acceleration in the absence of gravitational fields, then the same redshift must be observed by an experiment performed under conditions where there is a uniform gravitational field, but no acceleration. Consequently, by the principle of equivalence, one can derive equation (7.20) as applied to the gravitational situation.

§7.5. LOCAL FLATNESS, GLOBAL CURVATURE

The equivalence principle helps one to discern the nature of the spacetime curvature, whose existence was inferred from Schild's argument. Physics is the same in an accelerated frame as it is in a laboratory tied to the Earth's surface. Thus, an Earth-bound lab can be regarded as accelerating upward, with acceleration

g

, relative to the Lorentz frames in its neighborhood.* Equivalently, relative to the lab and the Earth's surface, all Lorentz frames must accelerate downward. But the downward (radial) direction is different at different latitudes and longitudes. Hence, local Lorentz frames, initially at rest with respect to each other but on opposite sides of the Earth, subsequently fall toward the center and go flying through each other. Clearly they cannot be meshed to form the single global Lorentz frame, tied to the

Earth, that was assumed in Schild's argument. This nonmeshing of local Lorentz frames, like the nonmeshing of local Cartesian coordinates on a curved 2 -surface, is a clear manifestation of spacetime curvature.
Geographers have similar problems when mapping the surface of the earth. Over small areas, a township or a county, it is easy to use a standard rectangular coordinate system. However, when two fairly large regions are mapped, each with one coordinate axis pointing north, then one finds that the edges of the maps overlap each other best if placed at a slight angle (spacetime analog: relative velocity of two local Lorentz frames meeting at center of Earth). It is much easier to start from a picture of a spherical globe, and then talk about how small flat maps might be used as good approximations to parts of it, than to start with a huge collection of small maps and try to piece them together to build up a picture of the globe. The exposition of the geometry of spacetime in this book will therefore take the first approach. Now that one recognizes that the problem is to fit together local, flat spacetime descriptions of physics into an over-all view of the universe, one should be happy to jump, in the next chapter, into a broadscale study of geometry. From this more advantageous viewpoint, one can then face the problem of discussing the relationship between the local inertial coordinate systems appropriate to two nearby regions that have slightly different gravitational fields.

There are actually two distinguishable ways in which geometry enters the theory of general relativity. One is the geometry of lengths and angles in four-dimensional spacetime, which is inherited from the metric structure

d s^{2}

of special relativity. Schild's argument already shows (without a direct appeal to the equivalence principle) that the special-relativistic ideas of length and angle must be modified. The modified ideas of metric structure lead to Riemannian geometry, which will be treated in Chapters 8 and 13. However, geometry also enters general relativity because of the equivalence principle. An equivalence principle can already be stated within Newtonian gravitational theory, in which no concepts of a spacetime metric enter, but only the Euclidean metric structure of three-dimensional space. The equivalence-principle view of Newtonian theory again insists that the local standard of reference be the freely falling particles. This requirement leads to the study of a spacetime geometry in which the curved world lines of freely falling particles are defined to be locally straight. They play the role in a curved spacetime geometry that straight lines play in flat spacetime. This "affine geometry" will be studied in Chapters 10-12. It leads to a quantitative formulation of the ideas of "covariant derivative" and "curvature" and even "curvature of Newtonian spacetime"!

Nonmeshing of local Lorentz frames motivates study of geometry

Two types of geometry relevant to spacetime:

Riemannian geometry (lengths and angles)

Affine geometry (" straight lines" and curvature)

PART
III

THE MATHEMATICS OF CURVED SPACETIME

Wherein the reader is exposed to the charms of a new temptress-
Modern Differential Geometry-and makes a decision: to embrace her for eight full chapters; or, having drunk his fill, to escape after one.

- Proper time equals Lorentz coordinate time for both observers, since they are at rest in the Earth's Lorentz frame.
*This upward acceleration of the laboratory, plus equation (6.18) for the line element in an accelerated coordinate system, explains the nonequality of the bottom and top edges of the parallelograms in Figure 7.1.

снартев 3

THE ELECTROMAGNETIC FIELD

§3.1. THE LORENTZ FORCE AND THE ELECTROMAGNETIC FIELD TENSOR

Box 3.1 LORENTZ FORCE LAW AS BOTH DEFINER OF FIELDS AND PREDICTER OF MOTIONS

EXERCISE

Exercise 3.1.

§3.2. TENSORS IN ALL GENERALITY

EXERCISES

Exercise 3.2. TRANSFORMATION LAW FOR COMPONENTS OF A TENSOR

Exercise 3.3. RAISING AND LOWERING INDICES

Exercise 3.4. TENSOR PRODUCT

Exercise 3.5. BASIS TENSORS

Box 3.2 THE METRIC IN DIFFERENT LANGUAGES

A. Geometric Language

B. Component Language

C. Coordinate-Based Geometric Language

D. Connection to the Elementary Concept of Line Element

Exercise 3.6. Faraday MACHINERY AT WORK

§3.3. THREE-PLUS-ONE VIEW VERSUS GEOMETRIC VIEW

§3.4. MAXVELL'S EQUATIONS

Exercise 3.7. MAXWELL'S EQUATIONS

EXERCISE

§3.5 WORKING WITH TENSORS

EXERCISES

Exercise 3.8. CONTRACTION IS FRAME-INDEPENDENT

Exercise 3.9. DIFFERENTIATION

Exercise 3.10. MORE DIFFERENTIATION

Box 3.3 TECHNIQUES OF INDEX GYMNASTICS

Equation

Exercise 3.11. SYMMETRIES

Exercise 3.12. SYMMETRIZATION AND ANTISYMMETRIZATION

Exercise 3.13. LEVI-CIVITA TENSOR

Exercise 3.14. DUALS

Exercise 3.15. GEOMETRIC VERSIONS OF MAXWELL EQUATIONS

Exercise 3.16. CHARGE CONSERVATION

Exercise 3.17. VECTOR POTENTIAL

Exercise 3.18. DIVERGENCE OF ELECTROMAGNETIC STRESS-ENERGY TENSOR

ELECTROMAGNETISM AND DIFFERENTIAL FORMS

§4.1. EXTERIOR CALCULUS

Box 4.1 DIFFERENTIAL FORMS AND EXTERIOR CALCULUS IN BRIEF

A. Algebra I (applicable to any vector space)

An application

Two applications

Applications:

Box 4.1 (continued)

Two applications

An application

Box 4.1 (continued)

D. Algebra II (applicable to any vector space with metric)

Applications

Box 4.1 (continued)

§4.2. ELECTROMAGNETIC 2-FORM AND LORENTZ FORCE

Box 4.2 ABSTRACTING A 2-FORM FROM THE CONCEPT OF "HONEYCOMBLIKE STRUCTURE," IN 3-SPACE AND IN SPACETIME

§4.3. FORMS ILLUMINATE ELECTROMAGNETISM, AND ELECTROMAGNETISM ILLUMINATES FORMS

Box 4.3 DUALITY OF 2-FORMS IN SPACETIME

Duality Rotations

§4.4. RADIATION FIELDS

§4.5. MAXWELL'S EQUATIONS

§4.6. EXTERIOR DERIVATIVE AND CLOSED FORMS

Box 4.4 THE PROGRESSION OF FORMS AND EXTERIOR DERIVATIVES

From Scalar to 1-Form

General 1-Form β = β α d x α β = β α d x α beta=beta_(alpha)dx^(alpha)\boldsymbol{\beta}=\beta_{\alpha} \boldsymbol{d} x^{\alpha}β=βαdxα

O-Form to 1 1 1\mathbf{1}1-Form to 2 -Form? No!

From 2-Form to 3-Form to 4-Form? No!

From 4-Form to 5-Form? No!

Results of Exterior Differentiation, Summarized

New Forms from Old by Taking Dual (see exercise 3.14)

Duality Plus Exterior Differentiation

§4.7. DISTANT ACTION FROM LOCAL LAW

EXERCISES

Exercise 4.1. GENERIC LOCAL ELECTROMAGNETIC FIELD EXPRESSED IN SIMPLEST FORM

Exercise 4.2. FREEDOM OF CHOICE OF 1-FORMS IN CANONICAL representation of generic local field

Exercise 4.3. A CLOSED OR CURL-FREE 1-FORM IS A GRADIENT

Exercise 4.4. CANONICAL EXPRESSION FOR A ROTATION-FREE 1-FORM

Exercise 4.5. FORMS ENDOWED WITH POLAR SINGULARITIES

Exercise 4.6. THE FIELD OF THE OSCILLATING DIPOLE

Exercise 4.7. THE 2-FORM MACHINERY TRANSLATED INTO TENSOR MACHINERY

Exercise 4.8. PANCAKING THE COULOMB FIELD

Exercise 4.9. COMPUTATION OF SURFACE INTEGRALS

Exercise 4.10. WHITAKER'S CALUMOID, OR, THE LIFE OF A LOOP

General 1-Form $β = β_{α} d x^{α}$

O-Form to $1$ -Form to 2 -Form? No!

CHAPTER
5

D. 3-Volume Swept Out in Time $Δ τ$ by Two-Dimensional Top of a Soap Box (Example)

§5.10. SAMPLE APPLICATIONS OF $\nabla \cdot T = 0$